ࡱ>  {t{{q` ebjbjqPqP .8::BN'1 8  d >#"`#`#$ua$q ]xaaaaaaafh]iDa;[:uaa `#$6dFFF^ `#6$aFaFF(pV:/$  Om[H,^,`?\"d0d,|i: i:/R/6 i;Ֆ FCՖՖՖaa^ՖՖՖd\  Chng I: Tng quan v c s d liu 1. Mt s khi nim 1.1. C s d liu C s d liu (Database, vit tt l CSDL) l mt lnh vc nghin cu cc m hnh, nguyn l, phng php t chc d liu trn cc vt mang tin. d dng cho vic gii thch cc khi nim, trc ht xem xt h thng bn v my bay bng my tnh. D liu lu tr trong my tnh bao gm thng tin v hnh khch, chuyn bay, ng bay v...v. Mi thng tin v mi quan h ny c biu din trong my thng qua vic t ch ca khch hng. Vy lm th no biu din c d liu v m bo cho hnh khch i ng chuyn. D liu nu trn c lu trong my theo mt quy nh no v c gi l c s d liu. Nh vy, CSDL l tp hp cc thng tin c quan h vi nhau c lu tr trong my tnh theo mt qui nh nht nh nhm phc v cho mt mc ch no . 1.2 H qun tr c s d liu (Database Management System- HQTCSDL) H qun tr c s d liu l h thng cc chng trnh nhm to lp, x l, thay i, qun l v khai thc cc CSDL. Mt s HQTCSDL thng dng hin nay: Foxpro, Access, Oracle,..vi cc phin bn khc nhau. C hai c im phn bit mt HQTCSDL vi cc h thng lp trnh khc: Kh nng qun l nhng d liu c nh. Kh nng truy xut c hiu qu mt s lng ln d liu. im (1) khng nh rng c mt CSDL tn ti thng xuyn v ni dung ca CSDL ny l nhng d liu c HQTCSDL truy xut v qun l. im (2) phn bit mt HQTCSDL vi cc h thng qun l tp tin cng qun l d liu c nh nhng ni chung khng cho php truy xut nhanh chng cc thnh phn tu ca d liu. Ngoi ra cn c mt s kh nng khc thng gp trong cc HQTCSDL trn th trng. HQTCSDL h tr t nht mt m hnh d liu nh ngi s dng c th xem c d liu. HQTCSDL h tr mt s ngn ng bc cao cho php ngi s dng nh ngha cc cu trc d liu, truy xut d liu v thao tc d liu. HQTCSDL qun l cc giao dch, ngha l cho php nhiu ngi s dng truy xut ng thi v chnh xc n mt CSDL. HQTCSDL iu khin qu trnh truy xut, l kh nng gii hn cc qu trnh truy xut d liu ca nhng ngi khng c php v kh nng kim tra tin cy ca d liu. HQTCSDL c kh nng t thch ng l kh nng phc hi li d liu do s c ca h thng m lm mt d liu. 2. Cc m hnh d liu M hnh CSDL( Database Model ) M hnh CSDL l mt h hnh thc ton hc gm c hai phn: Mt h thng k hiu m t d liu. Mt tp hp cc php ton thao tc trn d liu . Mt s m hnh CSDL thng dng 2.1. M hnh mng Khi nim: M hnh mng l mt m hnh s thc th lin kt vi tt c cc lin kt c hn ch l lin kt hai ngi nhiu  mt. Hn ch ny cho php chng ta s dng  th c hng n gin  biu din d liu trong m hnh ny. Trong m hnh mng, cc tp thc th c chuyn thnh cc kiu bn ghi logic. Cc kiu bn ghi logic bao gm mt tp cc trng, mi trng cha gi tr l mt s nguyn hay mt xu k t& Tp tn cc trng v cc kiu ca chng cu thnh quy cch bn ghi logic. 2.2. M hnh phn cp Mt m hnh phn cp n gin l mt m hnh mng m l mt rng (tp cc cy) trong c tt c cc mc ni tr theo hng t con n cha. Chng ta s tip tc s dng cc thut ng ca m hnh mng: kiu bn ghi logic khi chng ta ni v m hnh phn cp. 2.3. M hnh quan h : L m hnh da vo k hiu l tp cc tn v c s ton hc ca n l cc php ton tp hp v nh x. N l m hnh ph bin hin nay. Tp cc php ton trong m hnh ny da trn hai h k hiu: h k hiu i s v h k hiu logic. 2.4. M hnh thc th lin kt : L m hnh cho php m t cc thc th thng qua cc thuc tnh v mi lin h gia cc thc th. Mt trong cc cch biu th m hnh thc th l dng th, s khi. Mc ch ca m hnh thc th- lin h l cho php m t lc khi nim ca mt t chc m khng cn ch n tnh hiu qu hoc thit k CSDL vt l c mong i nh phn ln cc m hnh khc. Ngi ta tha nhn rng S thc th lin kt ( Entity Relationship Diagram) c th chuyn v lc khi nim cc m hnh khc (v d m hnh quan h) m trn cc h thng CSDL thc s c xy dng mt cch kh n gin. 2.4.1 Thc th Thut ng thc th (Entity) khng c mt nh ngha hnh thc. Thc th l mt s vt tn ti v phn bit c, ngha l c th phn bit c thc th ny vi thc th khc. V d mi con ngi l mt thc th, mi chic xe my l mt thc th. Khi nim v Tnh phn bit c rt gn vi c tnh nhn dng i tng v th m hnh thc th lin h c xem nh l m hnh hng i tng. 2.4.2. Tp thc th Mt nhm bao gm tt c cc thc th tng t to ra mt tp thc th. V d 1.4: Cc tp thc th + Tt c mi ngi trong mt c quan. + Tt c mi ngi c tc . + Tt c mi xe gn my. Mt trong cc bc mu cht chn mt lc cho th gii thc khi n thuc v mt CSDL c th l la chn cc tp thc th. Khi nim tp thc th l mt khi nim mc lc . Khi nim mc th hin tng ng l tp con hin hnh ca tt c cc phn t ca mt tp thc th cho trc ang hin din trong CSDL. 2.4.3. Thuc tnh v kho + Thuc tnh: Cc c tnh ca tp thc th gi l cc thuc tnh. Mi thuc tnh ca tp thc th ly gi tr trn mt min dnh cho thuc tnh . Thng th min gi tr i vi mi thuc tnh l mt tp s nguyn, tp cc s thc hoc chui k t nhng cng khng loi tr cc kiu gi tr khc. Th d mt tp thc th con ngi c th khai bo c cc thuc tnh nh h v tn (chui k t), chiu cao (s thc), ngy sinh (ngy thng nm),... Chn thuc tnh thch hp cho cc tp thc th l mt bc quan trng trong vic thit k lc CSDL khi nim. + Kha: Mi thuc tnh hoc mt tp cc thuc tnh dng xc nh mt cch duy nht mi thc th trong mt tp thc th gi l kha i vi tp thc th . V nguyn tc, mi thc th c mt kha, bi v mi thc th u c th phn bit c vi thc th khc. Nu khng chn c mt tp cc thuc tnh c cha mt kha cho mt tp thc th th khng c kh nng phn bit c thc th ny vi thc th kia trong tp thc th . Trong trng hp ny th cc s m thng c gn lm thuc tnh kha. V d 1.5: Mt tp thc th ch bao gm cc cng dn Vit Nam c th dng thuc tnh S chng minh th (IDNO) lm kho. Tuy nhin, nu mun xc nh mt cch duy nht cc cng dn ca nhiu quc gia th khng th m bo c hai quc gia s khng dng hai s chng minh th ging nhau. V vy, mt kho thch hp phi gm mt cp thuc tnh IDNO v COUNTRY (nc). 2.4.4. Mi lin h Mi lin h (Relationship) gia cc tp thc th l mt danh sch c th t ca cc tp thc th. Mt tp thc th c bit c th xut hin nhiu ln trong danh sch. Danh sch cc tp thc th ny l mt khi nim mc lc ca mt mi lin h. Nu c mt mi lin h R gia cc tp thc th E1, E2, E3, ..., EK th th hin ca R l mt tp cc k b, mt tp nh th c gi l mt tp lin h. Mi k b (e1,. .., ek) trong tp lin h R khng nh rng cc thc th e1, e2,,ek trong ei ( Ei ; i =1, , k lin kt vi nhau thnh mt nhm trong mi lin h R. Trng hp thng dng nht l k =2. V d 1.7: C mt tp thc th PERSONS c mi lin h MOTHER_OF v danh sch cc thc th ca n l PERSONS, PERSONS. Tp lin h tng ng vi mi lin h MOTHER_OF gm tt c nhng cp (P1, P2) sao cho c th P2 l m c th P1. Mt cch khc biu din thng tin ny l tha nhn tn ti tp thc th MOTHERS v mi lin h MOTHERS isa PERSONS. Do vy mi lin h MOTHER_OF l danh sch cc tp PERSONS, MOTHERS 2.4.5. Thuc tnh kho vay mn Trong mi lin h ISA nu A isa B th ng nhin kho ca A s l kho ca B v nhng thuc tnh ny khng xut hin nh thuc tnh ca tp A m vay mn ca tp B. V d 1.8: (Trong v d 1.6) Kho ca tp DANGVIEN s l thuc tnh MANV ca NHANVIEN. V vy, mt ng vin c xc nh mt cch duy nht nh vo MaNV ca NHANVIEN. Nhiu khi thuc tnh kho ca tp thc th A l mt thuc tnh ca tp thc th B nh mi lin h R khng phi ISA. iu ny cn thit cung cp cho mi thc th a trong A mt thc th b trong B. V d 1.9: Chng hn, (trong v d 1.5) mi cng dn u c thuc tnh Country v cng vi IDNO to ra mt kho cho mi cng dn iu c ngha l vic thit k CSDL xem cc quc gia nh mt kiu thc th khc v c mt mi lin h Country_of lin kt gia cc cng dn vi cc quc gia. V vy, cc cng dn phi mn thuc tnh Country cng vi IDNo lm kho. Trong cc v d trn thuc tnh MANV, Country l cc thuc tnh kho vay mn. 2.4.6. S thc th lin kt Qui c: Cc hnh ch nht biu din cc tp thc th. Cc vng trn hoc elp biu din cc thuc tnh. Chng c lin kt vi cc tp thc th bng cc cnh ( v hng). Cc thuc tnh l cc thnh phn ca mt kha cho mt tp thc th s c gch di. Trng hp c bit nu mt tp thc th ch c mt thuc tnh th c th gi tp thc th bng tn thuc tnh ca tp. Khi tp thc th s l mt vng trn ch khng phi l hnh ch nht v n gn kt vi cc mi lin h m tp hm cha.  Cc hnh thoi biu din cc mi lin h. Chng c lin kt vi cc cc tp thnh vin bi cc cnh v hng hoc c hng (cc cung). V d 1.10: Gi s ta c 3 tp thc th EMPS (Nhn vin) c cc thuc tnh ENO, EName, Salary ( ENO l kho), DEPTS (Phng) c cc thuc tnh DNO, DName, Location (DnO l kho), MANAGERS (Trng phng) c duy nht thuc tnh ENO (ENO l kho). Hai tp u lin kt nh mi lin h ASSIGNED_TO (thuc phng) v hai tp sau nh lin h MANAGES (qun l). 2.4.7.Tnh cht hm ca mi lin h m hnh ha y th gii thc cn phi phn loi cc mi lin h theo s lng cc thc th t mi tp tham gia vo trong mi lin h. a) Mi lin h mt mt Mt mi lin h mt - mt (one to one ralationship) l mt mi lin h m vi mi thc th trong mt tp thc th ny ch c nhiu nht mt phn t c lin kt trong tp thc th kia. V d 1.11: Mi lin h MANAGES gia depts v MANAGER trong v d trn c khai bo l mi lin h mt - mt. Nu nh vy trong CSDL khng bao gi tm c nhiu trng phng cho mt phng v cng khng c ngi no qun l nhiu phng. C th l ti mt thi im mt phng no khng c trng phng, thm ch cng c th c ngi c tn trong danh sch trng phng li khng qun l mt phng no c. Tnh mt- mt ca mi lin h ny ch l mt gi thit v th gii thc. V vy, ngi thit k CSDL c th tu chn la. Vn c th cho rng mt ngi lnh o 2 phng thm ch 1 phng c 2 trng phng khi mi quan h trn l nhiu - nhiu. Tuy nhin nu cho MANAGES l mi lin h mt - mt s c ch hn khi thit k CSDL vt l. b) Mi lin h nhiu mt Mt mi lin h gia cc tp thc th E1 v E2 c gi l mi lin h nhiu mt (many - one relationship) nu mi thc th trong tp E2 c th khng lin kt vi thc th no hoc lin kt vi mt hay nhiu thc th trong tp E1, nhng mi thc th trong tp thc th E1 ch lin kt nhiu nht vi mt thc th trong tp thc th E2. V d 1.12: Mi lin h ASSGNED_TO gia EMPS v DEPS v d trn l mi lin h nhiu-mt, c ngha l mi nhn vin ch lm vic trong mt phng, mt vi nhn vin nh gim c chng hn khng gn cho mt phng no c v mt phng c nhiu nhn vin. Khi nim lin h nhiu - mt tng qut ha thnh mi lin h gia ba tp tr ln. Nu c mt mi lin h R gia cc tp E1, E2,..., Ek v vi cc thc th trong tt c cc tp thc th tr Ei ch c nhiu nht mt thc th ca Ei c lin h vi chng th ta gi R l mi lin h nhiu mt t E1, E2, ..., Ek (tr Ei) n Ei. c) Mi lin h nhiu nhiu Chng ta cng gp mi lin h nhiu nhiu, khng c mt hn ch no trn tp k b ca cc thc th khi xut hin trong tp lin h. Mt mi lin nhiu - nhiu (many-many relationship) l mt mi lin h m vi mi thc th trong mt tp thc th ny c th khng lin kt vi thc th no hoc lin kt vi mt hay nhiu thc th trong tp thc th kia. V d 1.13: Mi lin h SUPPLIES gia tp thc th SUPPLIERS gm cc thuc tnh SNO, SNAME, SADDR v tp thc th PRODUCTS gm cc thuc tnh PNO, PNAME, COLOR, WEIGHT l mi lin h nhiu - nhiu. Trong thc hnh cc mi lin h nhiu - nhiu thng hay gp nn phi cn thn trong cch din t nhng mi lin h ny trong lc khi nim ca CSDL thc s (thit k thc th - lin h khng phi l lc khi nim, ng hn ch l bng phc tho v cn chuyn i cc tp thc th v cc mi lin h v m hnh d liu m DBMS h tr). Nhiu m hnh d liu khng cho php biu din trc tip cc mi lin h nhiu - nhiu, yu cu phi phn chng ra thnh cc mi lin h nhiu - mt. 2.5. M hnh hng i tng : L m hnh cung cp c tnh nhn dng i tng. Trong mi lp i tng c c trng bi hai yu t: a) Tp cc thuc tnh (properties) nhn dng i tng. b) Tp cc phng thc (methods) thao tc vi i tng. Cu hi v bi tp chng 1 Nu khi nim v CSDL. 1.2 Nu khi nim HQTCSDL, phn bit n vi cc h thng lp trnh khc. 1.3 Nu kin trc mt h CSDL v phn bit cc mc ca n. 1.4 Nu khi nim lc v th hin, cho v d. 1.5 Nu khi nim m hnh d liu. 1.6 Nu cc khi nim thc th v tp thc th. Cho v d. 1.7 Nu cc khi nim kho ca tp thc th v cho v d. 1.8 Nu cc mi lin h gia cc tp thc th v cho v d. Chng II: M hnh c s d liu quan h 1.Cc khi nim c bn 1.1.Thuc tnh(Attribute): Thuc tnh l mt tnh cht ring bit ca mt i tng cn c lu tr trong CSDL phc v cho vic khai thc d liu v i tng. V d 1.1: i tng KHOA ( tng ng vi loi thc th KHOA trong m hnh thc th kt hp ) c cc thuc tnh Ma-khoa, Ten-khoa. Loi thc th LOP-HOC c mt s thuc tnh Ma-lop, Ten-lop, Nien-khoa, So-hoc-vien. Loi thc th MON-HOC c mt s thuc tnh Ma-mon, Ten-mon. Chng hn vi bi ton qun l im thi ca sinh vin: vi i tng sinh vin ta cn ch n cc c trng ring nh h tn, ngy sinh, gii tnh, a ch, lp & cc c trng ny gi l thuc tnh Cc thuc tnh c phn bit qua tn gi v phi thuc vo 1 kiu d liu nht nh (s, chui, ngy thng, & ). Trong cng 1 i tng khng c c 2 thuc tnh cng tn. - Mi thuc tnh c xc nh trn 1 min gi tr gi l min thuc tnh K hiu DOM (<tn thuc tnh>) V d: DOM (h tn ) = {char(30)} DOM (tui ) = N V d 1.2: Thuc tnh ngy trong thng c kiu d liu l s nguyn, min gi tr ca n l t 1 n 31. Hoc im thi ca sinh vin c min gi tr t 0 n 10 Ngi ta dng cc ch ci hoa A, B, C,&  biu din cc thuc tnh, hoc A1, A2,&  biu din mt s lng ln cc thuc tnh. 1.2. Quan h Min (domain) l mt tp cc gi tr, chng khc g mt kiu d liu (data type). V d tp cc s nguyn l mt min, tp cc xu k t to thnh tn ngi trong ting Anh c di khng qu 30 k t l mt min, tp hai s{0,1} cng l mt min v.v. Tch - Cc: Gi D1, D2, ..., Dn l cc min. Tch - Cc ca n min ny k hiu l D1xD2x...xDn l tp tt c n- b (n tuples) (v1, v2,. .., vn) sao cho vi thuc Di, vi i = 1...n. V d 2.1: n=2, D1={0,1}, D2={a,b,c}. Khi : D1 x D2={(0,a),(0,b),(0,c),(1,a)(1,b),(1,c)}. Quan h (Relation) l mt tp con ca tch - Cc ca mt hoc nhiu min. Nh vy, mi quan h c th l v hn. y lun gi thit rng, quan h l mt tp hu hn. V d 2.2: {(0,a),(0,c),(1,a)(1,b)} l mt quan h l tp con ca tch cc D1xD2 c cp trn. Tp rng ( cng l mt quan h. Mi phn t ca quan h gi l mt b (tuples). Quan h n - ngi l tp con ca tch cc D1 x D2 x. .. x Dn ca n - min. Khi mi b ca quan h c n thnh phn (v1, v2, ..., vn) c gi l mt n-b. 1.3. B gi tr (Tuple) B l tp hp mi gi tr lin quan n tt c cc thuc tnh ca 1 lc  quan h Ngi ta thng dng cc ch ci thng (t, p, q,& )  biu din cc b. Nu b t thuc quan h r th: t  EMBED Equation.DSMT4  r V trc quan th mi quan h xem nh 1 bng, trong  mi ct l thng tin v 1 thuc tnh v mi dng l thng tin v mt b. MASVHOTENNGAYSINHMALOPDIACHIHOCBONG1462Nguyn vn B15/03/1990CTIN1Hi Dng1200001390Trn Vn A03/07/1989CTIN2Nam nh120000 1.4.Lc quan h Tp cc tn thuc tnh cho mt quan h c gi l mt lc quan h ( Relation Scheme). Nu chng ta t tn cho mt quan h l R v lc quan h ca n c cc thuc tnh A1, A2,..., An th lc quan h ny c th vit di dng R(A1, A2,..., An). Nh vy khi ta ni cho mt lc quan h R(A1, A2,..., An) c ngha l ta cho mt tp thuc tnh A1, A2,..., An v trn tn ti mt quan h R. V d 2.2: Hnh 2.3 cho thy mt quan h NHAN_VIEN c cc thuc tnh HO_TEN, NAM_SINH, NOI_LAM_VIEC v LUONG l mt quan h 4 ngi. NHAN_VIEN (Ho_Ten Nam_Sinh Noi_lam_viec Luong)  T1 L Vn A 1960 Vin KHVN 425 T2 Hong Th B 1970 Trng HBK 390 T3 L Vn Sn 1945 Vin KHVN 425 Hnh 2.3 Quan h Nhan_vien t1= (L Vn A 1960, Vin KHVN, 425) l mt b ca quan h NHAN_VIEN. C th vit lc quan h ny l NHAN_VIEN (Ho_Ten, Nam_Sinh, Noi_lam_viec, Luong). 1.5. Th hin ca quan h Th hin (hoc cn gi l tnh trng) ca quan h R, k hiu bi TR, l tp hp cc b gi tr ca quan h R vo mt thi im. Ti nhng thi im khc nhau th quan h s c nhng th hin khc nhau. Th hin (hay tnh trng) ca cc lc quan h con TRi gi l tnh trng ca lc c s d liu . V d: Cc th hin ca quan h lp-hc v mn-hc Quan h lp-hc: M-lpTn-lpNin-khoS-HvinM-khoaQTKD1Qun tr kinh doanh QT0196-99154QTKDTCKT1Ti chnh k ton KT397-2000200TCKTCNTT1Cng ngh thng tin K198-2001120CNTT Quan h mn-hc: M-mnTn-mnS-v-hc-trnhTCKTTi chnh k ton4CSDLC s d liu5KTCTKinh t chnh tr4LTCBLp trnh cn bn C5 1.6. Kho - Siu kho - Kho ch nh kho chnh kho ngoi Kho (key): Kho ca mt quan h R(A1, A2,...,An) l tp con ( ( K ( { A1, A2,...,An }, tho mn cc tnh cht sau y: 1) Vi bt k 2 b t1, t2 ( R u tn ti mt thuc tnh A(K sao cho t1[A] ( t2[A]. Ni mt cch khc. khng tn ti 2 b m c gi tr bng nhau trn mi thuc tnh ca K. iu kin ny c th vit t1[K] ( t2[K]. Do vy mi gi tr ca K l xc nh duy nht. 2) Khng c tp con thc s no ca K c tnh cht (1) V d 2.8: HANG_HOA ( MSMH TEN_HANG SO_LUONG )  10101 St phi 6 1000 10102 St phi 8 2000 20001 Xi mng 1000 Quan h HANG_HOA Trong v d trn biu din quan h HANG_HOA trong m s mt hng (MSMH) l kho. Mi gi tr MSMH u xc nh duy nht mt loi mt hng trong quan h HANG_HOA. Mt iu quan trng cn phi nh l kho ph thuc vo lc quan h khng ph thuc vo th hin ca quan h. Mt quan h c th c nhiu kho. Siu kha (Super key ): k l 1 kha ca quan h r(U) th mi tp hp k cha k u l kha ca quan h r(U)! k gi l siu kha ca quan h - Siu kha cha t thuc tnh nht c gi l kha ch nh. - Kha chnh: l kha c chn  ci t trong mt h qun tr c s d liu. Khi chn kha chnh ta phi ch cc tnh cht: Kha c tnh p dng khi n khng b st bt k trng hp no ca vn  Kha phi c tnh duy nht dng  phn bit b ny vi b kia trong quan h Kha c tnh nh nht khi ta b qua bt k thuc tnh no ca n th n khng cn tnh duy nht na Kha c tnh n nh khi gi tr ca kha khng thay i - Kha ngoi: mt thuc tnh c gi l kha ngoi nu n l thuc tnh ca mt lc  quan h ny nhng li l kha chnh ca lc  quan h khc 1.7.Ph thuc hm Quan h R c nh ngha trn tp thuc tnh U = {A1,A2, & ,An}.  EMBED Equation.3 l hai tp con ca tp thuc tnh U. Nu tn ti mt nh x f: X ! Y th ta ni rng X xc nh hm Y, hay Y ph thuc hm vo X v k hiu l X !Y. Chng ta s tm hiu k hn v ph thuc hm trong Chng IV. 1.8.Rng buc ton vn Rng buc ton vn (vit tt l RBTV) l mt quy tc nh ngha trn mt (hay nhiu) quan h do mi trng ng dng quy nh. l quy tc m bo tnh nht qun ca d liu trong CSDL. Mi RBTV c nh ngha bng mt thut ton trong CSDL. V d: Quan h CCVC(M-CBVC, H-tn, H-s-lng) Quy tc: H s lng ca cn b vin chc (CBVC) phi ln hn hay bng 1.00 v nh hn hay bng 10.00. Thut ton:  EMBED Equation.3  th cc.H-s-lng>=1 & cc.H-s-lng<=10. Cc khi nim cng nh vn ch yu ca RBTV s c trnh by chi tit trong Chng IV. 1.9.Cc thao tc c bn trn cc quan h Ba thao tc c bn trn mtj quan h, m nh CSDL c thay i, l Thm(Insert), Xo (Delete) v Sa (Update) cc b gi tr ca quan h. 1.9.1.Php thm (chn) mt b mi vo quan h: Php thm mt b vo quan h R = {A1, A2, & , An}: INSERT (R; A1=d1, A2 = d2, & , An = dn} trong  Ai (i = 1, 2, & , n) l tn cc thuc tnh v di ( dom (Ai). V d: Thm mt b t5 = (Vi Vn Vit, 1986, Caoangcodien, 3.36) vo quan h Nhanvien (Hoten, Namsinh, Noilamviec, Luong): INSERT (Nhanvien; Hoten = VuVanViet, Namsinh = 1986, Noilamviec = Caodangcodien, Luong = 3.36). Ch : - Nu xem th t ca cc trng l c nh, khi  ta c th biu din php chn mt cch ngn gn hn: INSERT (R; d1, d2, & , dn). Mc ch ca php chn l thm mt b vo mt quan h nht nh; kt qu ca php tnh ny c th gy ra mt s sai st vi nhng l do sau: + B mi thm vo khng ph hp vi lc  quan h cho trc. + Mt gi tr ca mt thuc tnh no  nm ngoi min gi tr ca thuc tnh . + Gi tr kho ca b mi c th l gi tr  c trong quan h ang lu tr. Do vy, ty tng h c th s c nhng cch khc phc ring. 1.9.2. Php loi b b khi quan h Php loi b (DELETE) l php xo mt b ra khi mt quan h cho trc. DELETE (R; d1, d2, & , dn). V d: khi cn loi b mt b, chng hn t2 t quan h Nhanvien: DELETE (Nhanvien; HaiHa; 1987, Khoatin, 4.45). Ch : khng phi lc no php loi b cing cn y  thng tin v c b cn loi. Khi ta c gi tr ca b  ti cc thuc tnh kho K = {B1, & , Bn}, lc  php loi b c dng: DELETE (R; B1 =e1, & , Bi = ei). 1.9.3. Php sa i gi tr ca cc thuc tnh ca quan H Khi cn iu chnh mt s gi tr no  ti mt s thuc tnh, ta s dng php thay i. Gi tp {C1, & , Cp}( {A1, & , An} l tp cc thuc tnh m ti  casc gi tr ca b cn thay i: CH (R; A1=d1, A1 = d2,& , An =d2; C1=e1, C2 = e2,& , Cp=ep} Nu K = {B1, & , Bm} l kho ca quan h, khi  ch cn vit: CH (R; B1=b1, & , Bm=bm; C1=e1,& , Cp=ep). V d: khi cn thay i lng ca ng H trong quan h Nhanvien ta vit: CH (Nhanvien; Hoten =  Ha , Luong = 4.56). 2.Cc php ton trn i s tp hp 2.1.Php hp Hp ca hai quan h R v S l mt quan h, k hiu l R(S v l tp tt c cc b t sao cho t(R hoc t(S. Biu din hnh thc php hp c dng : R ( S = { t / t ( R hoc t (S }. Ta ch c th dng php hp cho cc quan h cng ngi, v vy tt c cc b trong kt qu s c cng s lng thnh phn. Tn thuc tnh trong cc quan h s b b qua khi thc hin php hp v quan h thu c c th gn cc thuc tnh ty , th t ca cc thuc tnh trong cc quan h phi c tn trng. Cc iu ny tng t i vi cc php ton khc nh hiu, giao, tch Descartes. V d 2.1: (a) R (A B C )  a1 b1 c1 a1 b2 c1 a2 b2 c2  S (A B C)  a2 b1 c2 a2 b2 c2  R (S (A B C)  a1 b1 c1 a1 b2 c1 a2 b2 c2 a2 b1 c2 (b)  R (A B C ) a b c d a f c b d S ( D E F) b g a d a f  R (S (A B C) a b c d a f c b d b g aCh : Trong v d trn, ta thy hai quan h R v S cng ngi, ta vn c th ly hp ca chng d cc ct ca hai quan h trn mang tn khc nhau, min l cc quan h c cng s lng cc thnh phn. Tuy vy quan h thu c s khng c tn r rng cho cc ct (ta c th khng vit tn ct hoc t tn cho cc ct kt qu l A, B, C nhng n mang ngha mi) 2.2. Php giao Giao ca hai quan h R v S l mt quan h, k hiu l R(S v l tp tt c cc b t sao cho t thuc c R v S. Biu din hnh thc php giao c dng R ( S = (t / t(R v t( S(. V d 2.2: R( A B C ) a1 b1 c1 a1 b2 c1S( D E F ) d1 e1 f1 a1 b1 c1 d1 e2 f2 R ( S (A B C) a1 b1 c1  2.3.Php tr Hiu ca hai quan h R v S kh hp l mt quan h k hiu l R-S v l tp tt c cc b t sao cho t thuc R nhng khng thuc S. Biu din hnh thc php c dng : R - S =( t / t(R v t( S(. V d 2.3: * Vi R, S v d 2.1 (a) trn:  R-S ( A B C ) a1 b1 c1 a1 b2 c1 * Vi R, S v d 2.1 (b) trn: R-S ( A B C )  a b c c b d 2.4.Php tch cc (Descartes) R l quan h n - ngi v S l quan h m - ngi. Tch cc ca hai quan h R v S k hiu l RxS l tp tt c cc (n+m) - b vi n thnh phn u l mt b thuc R v m thnh phn sau l ca mt b thuc S. Biu din hnh thc c dng RxS ={t / t c dng ( a1, a2,..., an, b1, b2,..., bm ), trong ( a1,..., an) (R v (b1,..., bm )(S}. V d 2.4: R( A B C ) a1 b1 c1 a1 b2 c1S( D E F ) d1 e1 f1 d2 e2 f1 d1 e2 f2 RxS ( A B C D E F ) a1 b1 c1 d1 e1 f1 a1 b1 c1 d2 e2 f1 a1 b1 c1 d1 e2 f2 a1 b2 c1 d1 e1 f1 a1 b2 c1 d2 e2 f1 a1 b2 c1 d1 e2 f2 R(A B C ) a b c d a f c b d S ( A B C ) a b c d a f  RxS (A B C D E F ) a b c a b c a b c d a f d a f a b c d a f d a f c b d a b c c b d d a f 2.5.Php chia - Trong quan h r xc nh trn quan h U1 l quan h bc n - Trong quan h s xc nh trn quan h U2 l quan h bc m Vi n > m Khi  php chia ca r cho s l 1 tp cc b t c bc l n-m sao cho mi b gi tr u ( s th b ghp c gia (t,u) ( r C php: r ( s = { t | ( u ( s ( (t,u) ( r} v d 2.5: R ( A B C D ) S (C D ) R(S= (A B ) a b c d c d a b a b e f e f c d b c e f c d c d c d e f a b d e 3.Cc php ton trn i s quan h 3.1.Php chiu (( Projection ) Cho quan h R(A1, A2,...,An), X( { A1, A2,...,An }. Php chiu quan h R trn tp thuc tnh X l mt quan h trn tp thuc tnh X, k hiu l (X (R) v c biu din hnh thc l: (X (R)= ( t[X] ( t ( R(. Trong t[X] l gi tr ca b t trn tp thuc tnh X. V d 2.18: R= { A,B,C,D }, X={ A,B };Y={ A,C} R ( A B C D ) (X(R) (A B ) (Y(R) (A C ) a1 b1 c1 d1 a1 b1 a1 c1 a1 b1 c1 d2 a2 b2 a2 c2 a2 b2 c2 d2 a2 c3 a2 b2 c3 d3 3.2. Php chn ( Selection ) Php chn l php tnh xy dng mt tp con cc b ca quan h cho, tho mn biu thc F xc nh. Biu thc F c din t bng mt t hp Boolean ca cc ton hng, mi ton hng l mt php so snh n gin gia hai bin l hai thuc tnh hoc gia mt bin l mt thuc tnh v mt hng, cho gi tr ng hoc sai i vi mi b cho khi kim tra ring b y. Cc php so snh trong biu thc F l <, =, >, >=, <= v (; Cc php logic l ( (v), ( (hoc) v ( (khng). Cho quan h R(A1, A2,...,An). Php chn quan h R vi iu kin F l mt quan h trn tp thuc tnh (A1, A2,...,An) k hiu l (F(R) Hnh thc ho php chn c nh ngha nh sau (F(R) = ( t (R ( F (t) = ng (. F(t) c hiu l gi tr ca biu thc F i vi b t V d 2.19: Cho quan h sau. R ( A B C D )  a1 b1 c1 d1 a1 b1 c1 d2 a2 b2 c2 d2 a2 b2 c2 d3 Cc php chn (A= a1 (R) = (A B C D )  a1 b1 c1 d1 a1 b1 c1 d2 ((A=a1) V (D =d2) (R) = ( A B C D )  a1 b1 c1 d1 a1 b1 c1 d2 a2 b2 c2 d2 3.3.Php kt ni ( Join ) Gi ( l mt trong cc php so snh ( =, >, >=, <, <=, ( (. Php kt ni ( ca quan h R i vi thuc tnh A v quan h S i vi thuc tnh B l nhng b t ca tch Descartes RxS sao cho t[A] ( t[B] R (( S = ( (t (t (RxS v t[A] ( t[B] (. A ( B c bit khi php snh ( l php = th php kt ni gi l kt ni bng. Trng hp kt ni bng i vi hai thuc tnh cng tn l A v kt qu thu c loi b i mt trong hai ct R.A hoc S.A, th php kt ni gi l kt ni t nhin v s dng k hiu ( thay cho ((. V d 2.21: R(A B C ) S( C D E ) (A B C F D E)  a1 1 1 1 d1 e1 a1 1 1 1 d1 e1 a2 2 1 2 d2 e2 a2 2 1 1 d1 e1 a1 2 2 3 d3 e3 a2 2 1 2 d2 e2 a1 2 2 1 d1 e1 a1 2 2 2 d2 e2 Kt qu ni t nhin :  (A B C D E ) a1 1 1 d1 e1 a2 2 1 d1 e1 a1 2 2 d2 e2 3.4. Cc php ton kt ni khc Mc ny trnh by 3 php ton kt ni m rng khc c bit quan trng, m bn cht ca chng vn l kt ni.Chng c ci t trong mt s h qun tr CSDL nh Microsoft Access, SQL Server, Oracle. Cc php kt ni l: Kt ni ni (Inner Join), kt ni trI (Left Join) v kt ni phi (Right Join). 3.4.1. php kt ni ni (Inner Join) Thc cht l php kt ni bng trnh by trn. Tuy nhin, ngay c trong trng hp hai thuc tnh so snh c cng tn th kt qu php kt ni vn gi li 2 tn thuc tnh . V d 3.4.1: Cho 2 quan h R(A B C) v S(A D E F) vi cc b gi tr nh di y. Kt qu php kt ni ni c cho trn bng pha bn phi. R(A B C ) S(A D E F) (A B C A D E F)  a1 b1 c1 a1 d1 e1 f1 a1 b1 c1 a1 d1 e1 f1 a2 b2 c2 a2 d2 e2 f2 a2 b2 c2 a2 d2 e2 f2 a3 b3 c3 a4 d4 e4 f4 a7 b7 c7 a7 d7 e7 f7 a5 b5 c5 a6 d6 e6 f6 a7 b7 c7 a7 d7 e7 f7 3.4.2. php kt ni tri (Left Join) Gi s c hai quan h R(A1 A2 An) v S(B1 B2 Bm). t=(a1,a2,, an) v u= (b1,b2,,bm) l hai b gi tr ca R v S. Gi r l b ghp ni u vo t (hay b gi tr t v u c xp cnh nhau) v k hiu l: r=(t,u)=(a1,a2,,an,b1,b2,,bm) B tNULL= (NULL,NULL,,NULL) l mt b c bit ca R gm n gi tr ca cc thuc tnh A1,A2,,An u l khng xc nh v uNULL=(NULL,NULL,,NULL) l mt b c bit ca S gm m gi tr ca cc thuc tnh B1,B2,,Bm u l khng xc nh.  EMBED Equation.3  v  EMBED Equation.3  l hai thuc tnh c th so snh c Php kt ni tri hai quan h R vi S trn cc thuc tnh A v B vi php so snh bng, vi gi thit l gi tr ct R[A] c th so snh tng ng c vi mi gi tr ca ct S[B], c nh ngha l: R (( S ={r =(t,u)|( EMBED Equation.3 v t.A ( u.B)hoc( EMBED Equation.3 ,u=uNULL vi t.A EMBED Equation.3 S[B])} A=B ngha l, tt c cc b r c c nh cch t b gi tr ca R v S xp cnh nhau, nu c gi tr ging nhau trn 2 thuc tnh kt ni v cc b r c c nh cch t b ca R vi cc b NULL ca S, nu khng tm c gi tr tng ng ca thuc tnh kt ni trn quan h S. V d 3.4.2: Vi quan h R v S cng cc b gi tr ca chng c cho trong v d 3.4.1, kt qu ca php kt ni tri ca R v S l: R (( S = Q(A B C A D E F)  R.A=S.A a1 b1 c1 a1 d1 e1 f1 a2 b2 c2 a2 d2 e2 f2 a3 b3 c3 - - - - a5 b5 c5 - - - - a7 b7 c7 a7 d7 e7 f7 K hiu du tr (-) trong cc thuc tnh ca S c hiu l gi tr khng xc nh (gi tr NULL) Cc dng c gi tr thuc tnh A ca R l a3 v a5 khng tm c gi tr ca thuc tnh A tng ng trong quan h S, nn phn cn li ca n c l khng xc nh. Qua bng kt qu trnh by trn, chng ta thy ngha ca php ton ny l nhm xc nh cc b gi tr ca quan h bn tri nhng khng c b tng ng trong quan h pha bn phi. 3.4.3. php kt ni phi (Right Join) Vn vi cc quan h R,S; Cc thuc tnh A, B; v cc b gi tr v,t,u, tNULL, uNULL c xc nh nh trn. Php kt ni phi hai quan h R vi S trn cc thuc tnh A v B vi php so snh bng, vi gi thit l gi tr ct R[A] c th so snh tng ng c vi mi gi tr ca ct S[B], c nh ngha l: R (( S ={r =(t,u)|( EMBED Equation.3 v t.A ( u.B)hoc( EMBED Equation.3 ,t=tNULLvi t.B EMBED Equation.3 R[A])} A=B ngha l, tt c cc b r c c nh cch t b gi tr ca R v S xp cnh nhau, nu c gi tr ging nhau trn 2 thuc tnh kt ni v cc b NULL ca R vi cc b ca S, nu khng tm c gi tr tng ng ca thuc tnh kt ni trn quan h R. V d 3.4.3: Vi quan h R v S cng cc b gi tr ca chng c cho trong v d 3.4.1, kt qu ca php kt ni trI ca R v S l: R (( S = Q(A B C A D E F)  R.A=S.A a1 b1 c1 a1 d1 e1 f1 a2 b2 c2 a2 d2 e2 f2 - - - a4 d4 e4 f4 - - - a6 d6 e6 f6 a7 b7 c7 a7 d7 e7 f7 K hiu du tr (-) trong cc thuc tnh ca R c hiu l gi tr khng xc nh (gi tr NULL) Cc dng c gi tr thuc tnh A ca S l a4 v a6 khng tm c gi tr ca thuc tnh A tng ng trong quan h R, nn phn u ca n c l khng xc nh. Qua bng kt qu trnh by trn, chng ta thy ngha ca php ton ny l nhm xc nh cc b gi tr ca quan h bn phi nhng khng c b tng ng trong quan h pha bn tri. Cu hi v bi tp chng 2 2.1 Nu nh ngha quan h, cho v d. 2.2 Nu cc php ton i s quan h, cho v d. 2.3 Cho hai quan h r v s nh sau: RABCD1000110011101111 SAEFG111122111110 Tnh R ( S, R ( S; R - S v S - R. Gi s X= (A, B(; Y= (A, C, D( Tnh ( X(R), ( X(S); ( X(R ( S), (Y (R ( S) Tnh ((A=1)( (D=0)(R); ((A=1)( (D=0)( R ( S) Tnh ((A=1)( (D=0) (( ABD(R*S)); ( ABD(((A=1)( (D=0) (R*S)); 2.4. Cho 3 quan h: RABCDa1b1c1d1a2b1c1d1a2b2c2d2a2b2c3d2 SABCDa1b1c3d1a2b1c1d1a1b1c1d1 UAEFGa1e2f1g1a2e1f1g1a1e2f2g2a4e2f2g2 Tm : a. R ( S, R ( S, R - S. b. R x U, R*U, ( A,B(R), ((A=a1)( (E= e2)(U). c. ( A,B (((A=a1)( (E= e2)(R*U)). d. ((A=a1)( (E= e2) ( ( A,B (R*U)). 2.5 Cho hai quan h R v S RABCDESDF000011100110101111100011Tnh R x S, R*S. 2.6 Cho hai quan h rABCDsAE10001110111011110001 Vi X = {A, B } a) Tnh: ( X (((A=1)( (E=0) (r * s)) b) Tnh: ((A=1)( (B=0) (( X (r * s)) c) Hy dng cc php ton i s quan h tr li cc cu hi sau. *Cho bit cc thng tin A, B, E c A=1 v E = 0 *Cho bit cc thng tin A, B, C c A=1 hoc E= 0 d) Tnh li kt qu ca cc biu thc cu c) 2.7 Cho 3 quan h: DS( Sbd, Hoten, Ngaysinh, Gioitinh, Quequan) SBD_PH( Sbd,Sophach) DTM( Sophach, Diem) Trong : Sbd: s bo danh Hoten: H v tn sinh vin Ngaysinh: Ngy sinh Gioitinh: Gii tnh Quequan: Qu qun Sophach: S phch Diem: im thi ca mn hc Hy dng cc php ton ca i s quan h tr li cc cu hi sau: Cho bit h v tn, ngy sinh, gii tnh, qu qun, im thi ca mi sinh vin. Cho bit h v tn, ngy sinh, gii tnh, im thi ca nhng sinh vin c im thi ( 5. Cho bit h v tn, ngy sinh, dim thi ca nhng sinh vin c gii tnh l Nu v qu Thai Binh. Cho bit h v tn, ngy sinh, im thi ca nhng sinh vin c qu Nam Dinh hoc Thai Binh v c im thi < 3 hoc im thi >8. 2.8 Cho 3 quan h: MatHang(Mamh, Tenmh, Mau, DVT) KhHang(Makh, Tenkh, Diachi, DT, Gioitinh) MuaBan(Mamh, Makh, Muaban, NgayMB, Soluong, Dongia) Trong : - Mamh: M mt hng Tenmh: Tn mt hng Mau: Mu mt hng DVT: n v tnh Makh: M khch hng Tenkh: Tn khch hng Diachi: a ch khch hng DT: in thoi Gioitinh: Gii tnh MuaBan: Mua bn trong khch hng mua th ghi l .F., khch hng bn th ghi l .T. NgayMB: Ngy mua hoc bn Soluong: S lng Dongia: n gi Hy dng cc php ton ca i s quan h tr li cc cu hi sau: Cho bit makh, tenkh, tenmh ca nhng khch hng bn mt hng c mamh = MH001 hoc mamh = MH002. Cho bit makh ca nhng khch hng bn mt hng mu vi s lng >100 trong qu I nm 2003. Cho bit tenkh, diachi, DT ca nhng khch hng bn mt hng mu Vang hoc xanh vi s lng >100. Cho bit tenkh, tenmh ca nhng khch hng c gii tnh l nam bn mt hng mu Den v mua mt hng mu Xanh vi 200> soluong>100. Cho bit makh cha tham gia mua bn ln no. Cho bit makh cha tham gia bn ln no. Cho bit makh, tenkh cha tham gia mua bn ln no. Cho bit makh mua mt hng mu xanh nhng khng bn mt hng mu . Cho bit makh mua mt hng mu xanh v bn mt hng mu . Cho bit makh mua mt hng mu xanh hoc bn mt hng mu . 2.9 Cho 2 quan h sau: Ds MaSVHotenNsGtQq001Nguyn Vn Anh20/05/1984NNam nh002Bi Thu Hu12/04/1987NThi Bnh003Nguyn Hi H12/10/1984NamH Nam004V Nh Qunh14/08/1985NNam nh005Trn Hi L23/08/1987NamThanh Ha Diem MaSVMonDiem001Ton8001L7003Ton9004Ton7003Ha5001Ha6003L9004L4004Ha10 Hy vit cc biu thc quan h thc hin cc cng vic sau v tnh cc kt qu ca cc biu thc : Cho xem MaSV, Hoten, Ns, Gt, Mon, Diem mi mn thi ca tng sinh vin. Cho xem MaSV, Hoten, Ns, Gt, Mon, Diem mn l ca cc hc sinh n. Cho xem Danh sch sinh vin khng d thi tt c cc mn. Cho xem Danh sch sinh vin khng d thi tt c cc mn gm cc thng tin MaSV, Hoten, Ns. Chng III Ngn ng d liu SQL 1. Khi qut v ngn ng d liu SQL Trong chng ny trnh by ngn ng d liu SQL (Structured Query Language). y l ngn ng con d liu quan h c xc nhn l rt mnh, ph dng v li d s dng. SQL c pht trin t ngn ng SEQUEL-2, th nghim v ci t ti trung tm nghin cu ca hng IBM San Jose, California cho h thng QTCSDL ln in hnh l System - R. trong System - R, SQL va ng vai tr l mt ngn ng nh ngha d liu - DDL va l ngn ng thao tc d liu - DML. SQL l mt ngn ng phi th tc, chun mc v in hnh. Do vy hin nay rt nhiu sn phm phn mm thng mi u c ci t SQL nh Oracle, Visual Foxpro, Visual Basic, Access.... Trong ti liu ny s trnh by cc kh nng ca ngn ng, ng thi cung cp cho bn c thm kinh nghim v cch nhn cc h QTCSDL tm gi l kinh in. Php ton c bn trong SQL l php nh x c miu t nh mt khi SELECT - FROM - WHERE. Cc mnh ca ngn ng SQL s c trnh by chi tit bng cc v d. Cc thut ng trong CSDL quan h nh quan h, thuc tnh, b,. .. c thay th bng cc thut ng nh bng (table), ct (column), bn ghi (record) hoc hng (row) ph hp vi ngha ca cc h mm ny. 2. Cc lnh lin quan n cu trc ca c s d liu 2.1 To bng C php: CR EATE TABLE ( [Not null ], [Not null ], [Not null ], PRIMARY KEY (Kho chnh), [UNIQUE (kho), ] [FOREIGN KEY (kho ngoi) REFERENCES Tn bng ] [Check iu kin rng buc, ]) Trong : - Tn bng: l xu k t bt k, khng trng t kho, khng cha du cch trng - Tn ct: l xu k t bt k, khng cha du cch trng. Trong mt bng tn ct l duy nht. Th t cc ct khng quan trng - Kiu d liu: Kiu d liu ca ct. Kiu d liu c th dng mt s loi d liu nh sau: Integer: S nguyn t -2 147 483 648 n 2 147 483 647 Smallinteger: S nguyn t -32 768 n 32 767 Number (n,p): S thp phn vi di ti a l n trong c p ch s phn thp phn. Float: S du phy ng Char(n): Xu k t c di ti a l n, n<=255 Varchar(n): kiu k t vi kch thc thay i t 0 n n k t. Gi tr ln nht ca n l 2000. Date: D liu dng ngy thng. Logical: D liu kiu logic 1 byte c gi tr hoc ng (True), hoc sai (False). - Ch th Notnull: ch rng ct khng nhn gi tr rng. Thuc tnh kho ngm nh l Notnull - Ch th PRIMARYKEY(kho chnh): Khai bo kho chnh ca bng - Ch th UNIQUE Kho: Khai bo cc kho khc nu c - Ch th FOREIGN KEY (kho ngoi) REFERENCES Tn bng : Khai bo cc kho ngoi ca bng - Ch th Check iu kin rng buc,& :Khai bo cc rng buc d liu. V d: To cc bng sau: SV(MaSV,TenSV,MaKh,iachi,Namsinh) KHOA(MaKh,TenKh,Vitri,TrKhoa) * To bng SV CR EATE TABLE SV(MaSV char (5) Notnull, TenSV varchar(20) Notnull MaKh char(2) Notnull iachi varchar(30) Notnull Namsinh Integer Notnull PRIMARYKEY(MaSV) FOREIGN KEY(MaKh) REFERENCES KHOA) *To bng KHOA: CR EATE TABLE KHOA(M aKh char(2) Notnull, T enKh varchar(20) Notnull Vitri varchar(30) Notnull TrKhoa varchar(20) Notnull PRIMARYKEY(MaKh) 2.2.Xo bng Mnh xo bng c dng tng qut nh sau: DROP TABLE tn_bng Bng c tn c ch ra trong mnh c xo khi CSDL. V d: Xo bng SV DROP TABLE SV 2.3. Thm, xo cc ct ca bng C php: * Thm ct ALTER TABLE Tn bng ADD Tn ct kiu d liu [kch thc] [Notnull ] * Xo ct ALTER TABLE Tn bng DROP Tn ct 3.Cc lnh cp nht c s d liu 3.1.Thm b vo bng Lnh chn d liu INSERT INTO c s dng chn mt dng hay hng d liu mi vo trong mt bng. C php: INSERT INTO tn bng VALUES (value1, value2,) Hoc c th ch r nhng ct c th mun chn d liu nh c php sau: INSERT INTO tnbng (Column1, column2,) VALUES (value1, value2) Chn mt dng mi: INSERT INTO Nhanvien VALUES ( 05 , Van ,  21/02/1987 ,  Giaovien ,  HaNoi , 2.400.000) Kt qu: MNVHotenNgaysinhNghenghiepDiachiLuong01Lan22/9/1988KetoanHaiPhong1.500.00002Nghia12/9/1988QuanlyHaiPhong1.500.00003Hai22/9/1989KetoanHaiPhong1.000.00004Trang30/1/1999TaivuBacNinh2.000.00005Van21/02/1987GiaovienHaNoi2.400.000V d chn d liu vo nhng ct ch nh c th: INSERT INTO Nhanvien (MNV, Hoten, Nghenghiep) VALUES ('Hoang', 'Baove') Kt qu: MNVHotenNgaysinhNghenghiepDiachiLuong01Lan22/9/1988KetoanHaiPhong1.500.00002Nghia12/9/1988QuanlyHaiPhong1.500.00003Hai22/9/1989KetoanHaiPhong1.000.00004Trang30/1/1999TaivuBacNinh2.000.00005HoangBaove 3.2.Cp nht ni dung ca b trong bng Lnh cp nht d liu c s dng sa i d liu trong mt bng. C php: UPDATE tn bng SET = , = , = [WHERE ]; V d: i tn Nghenghiep mt ngi c tn l Hoang thnh ngh Taivu UPDATE Nhanvien SET Nghenghiep = 'Taivu' WHERE Hoten= 'Hoang' Kt qu: MNVHotenNgaysinhNghenghiepDiachiLuong01Lan22/9/1988KetoanHaiPhong1.500.00002Nghia12/9/1988QuanlyHaiPhong1.500.00003Hai22/9/1989KetoanHaiPhong1.000.00004Trang30/1/1999TaivuBacNinh2.000.00005HoangTaivu Cp nht nhiu ct trong mt dng d liu Chng ta mun thay i a ch v thm vo tin lng: UPDATE Nhanvien SET Diachi = 'BacNinh', Luong = '3.000.000' WHERE Hoten = 'Hoang' Kt qu: MNVHotenNgaysinhNghenghiepDiachiLuong01Lan22/9/1988KetoanHaiPhong1.500.00002Nghia12/9/1988QuanlyHaiPhong1.500.00003Hai22/9/1989KetoanHaiPhong1.000.00004Trang30/1/1999TaivuBacNinh2.000.00005HoangTaivuBacNinh3.000.000 3.3. Xo cc b trong bng Lnh DELETE c s dng xo cc dng d liu mt bng. C php: DELETE FROM tnbng WHERE tnct = Gi tr V d: Xo mt hng: DELETE FROM Nhanvien WHERE MSV = '05' Kt qu: MNVHotenNgaysinhNghenghiepDiachiLuong01Lan22/9/1988KetoanHaiPhong1.500.00002Nghia12/9/1988QuanlyHaiPhong1.500.00003Hai22/9/1989KetoanHaiPhong1.000.00004Trang30/1/1999TaivuBacNinh2.000.000 Xo tt c cc hng C th xo tt c cc dng trong mt bng m khng cn phi xo bng. iuu ny c ngha l cu trc bng, ct v ch mc s khng b thay i v mt i. DELETE FROM tn bng Hoc DELETE * FROM tn bng 4. Cc lnh truy vn c s d liu Cu trc n gin nht trong SQL l khi SELECT c miu t v c php nh mt khi select -from -where Mt cch tng qut khi select bao gm 3 mnh chnh: Select: Xc nh ni dung ca cc ct cn a ra kt qu From: Xc nh cc bng cn ly thng tin ra Where: Xc nh cc bn ghi tho iu kin chn lc a ra kt qu. Ngoi ra, m rng kh nng ca ngn ng, khi select - from - where cn c b sung thm cc mnh group by, having, order by, cc hm mu v mt s phn mm cn thm c mnh compute, for browse. Trong cc phn sau s trnh by chi tit tng mnh . Dng tng qut ca khi lnh select c biu din nh sau: Select (( ( distinct ( danh_sch_ct / biu thc From danh_sch_tn_bng ( tn_cc_view ( where biu_thc_iu_kin( ( groupe by danh_sch_tn_ct ( (having biu_thc_iu_kin ( ( order by ( tn_ct ( s_th_t_ct ( biu_thc ( [ASC/ DESC ] ( Trong mnh where c biu din dng: WHERE [NOT] biu_thc php_snh biu_thc WHERE [NOT] tn_ct [ NOT ] LIKE xu_k_t WHERE [NOT] biu_thc [ NOT ] BETWEEN biu_thc and biu_thc WHERE [NOT] biu_thc [ NOT ] IN ({ danh_sch / cu_hi_con}) WHERE [NOT] EXISTS (cu_hi_con) WHERE [NOT] biu_thc php_snh { ANY ( ALL (cu_hi_con)} WHERE [NOT] tn_ct php_kt_ni tn_ct WHERE [NOT] biu_thc_logic WHERE [NOT] biu_thc_logic ( AND ( OR(( NOT ( biu_thc_logic 4.1. Tm thng tin t cc ct ca bng (php chiu) Trc ht lm quen vi cc cu hi ch lin quan ti mt bng. Trong mnh select c danh sch chiu. Danh sch ny xc nh tn cc ct cn c trong bng kt qu. Nu xut hin gi tr * c ngha l chn ton b cc ct ca bng. V d 4.1.1: Cho bit tt c cc thng tin v cc nh cung cp SELECT * FROM S Sau khi thc hin mt mnh SQL, bng kt qu ng l mt quan h (c ngha l khng c b trng nhau). Trong mnh select cn thm t kho Distinct. V d 4.1.2: a,Cho bit s hiu nhng mt hng c cung cp. SELECT DISTINCT PNo PROM SP b, Cho bit s hiu khch hng, s hiu mt hng, s tin m mi nh cung cp cung cp mt mt hng mi ln. SELECT SNO, Pno, QTY*PRiCE PROM SP 4.2.Chn cc b ca bng Mnh WHERE (php chn) V d 3.4: Tm s hiu nhng nh cung cp cung cp mt hng c s hiu l P2. SELECT DISTINCT SNo FROM SP WHERE PNO = P2 Trong SQL cc php snh c s dng bao gm >, <, >=, <=, = v <>. Cc php tnh trn dng cho mi loi d liu. X l xu k t gn ng cn gi l php tnh thng minh. Trong trng hp ngi s dng khng nh r tn ngi hoc a danh. ... V d 3.5: Cho bit tt c cc thng tin v cc nh cung cp c h l Phm. Khi c th vit: SELECT * FROM S WHERE SNAME LIKE Phm % Trong SQL s dng k hiu % l thay th cho mt xu con, du gch di _ thay th cho mt k t. A%B: xu k t bt k bt u bng ch A v kt thc bng ch B %A: xu k t bt k c k t kt thc bng l A A_B: xu gm 3 k t c k t th 2 l bt k A_: xu gm 2 k t c k t u l A. Ngoi cc php tnh thng thng SQL cn c th x l d liu dng ngy thng. V d 3.6: Tm s hiu mt hng, s lng ca nhng mt hng bn trc ngy 24 thng 4 nm 1994 l 10 ngy. SELECT PNo, QTY FROM SP WHERE {04/24/94} - SDATE = 10 Tm kim nh s dng IN v BETWEEN V d 3.7: Tm s hiu nhng mt hng cung cp c gi t 1000 n 2000. SELECT PNo FROM SP WHERE PRICE BETWEEN 1000 AND 2000 V d 3.8: Tm s hiu nhng nh cung cp cung cp t nht mt trong cc mt hng c s hiu P1, P2, P3. SELECT DISTINCT SNo FROM SP WHERE PNO IN ( P1, P2, P3 ) 4.3. Th t hin th cc bn ghi - Mnh ORDER BY V d 4.3.1: Tm tn, s hiu cc mt hng mu , kt qu a ra sp xp theo th t tng dn ca m s mt hng. SELECT PNAME, PNo FROM P WHERE COLOR = ORDER BY PNo ASC Ch : - ORDER BY dng sp xp kt qu d liu u ra. - Sau mnh ORDER BY l tn ct ri n chiu sp xp tng hoc gim (ASC hoc DESC ). Nu sp xp theo nhiu ct th th t sp xp u tin t tri qua phi, nu khng ch ra chiu sp xp th h thng ngm nh l ASC. - Biu thc sp xp sau mnh ORDER BY nu c mt trong mnh SELECT th c th dng th t xut hin ca n trong mnh trn thay cho biu thc sp xp. V d 4.3.2: a ra s hiu mt hng, tn mt hng c mu . Danh sch a ra sp xp theo chiu tng dn ca s hiu mt hng. SELECT PNAME, PNO FROM P WHERE COLOR = ORDER BY 2 4.4. Phn nhm d liu Mnh GROUP BY V d 4.4.1: Tm s hiu mt hng, s ln cung cp, tng s tin m cc nh cung cp cung cp mi mt hng. SELECT PNo, COUNT(PNO), SUM(QTY*PRICE) FROM SP GROUP BY PNo Trong mnh ny bng d liu SP c ly ra, sau phn thnh nhm theo s hiu mt hng (PNO) ri gp cc bn ghi ca nhm thnh mt bn ghi a ra kt qu. Ch : - Nu Group by theo nhiu ct th gia cc ct phn cch nhau bi du phy. - Tn cc ct c mt trong mnh Group by phi c mt trong mnh Select. 4.5.iu kin hin th cc bn ghi - Mnh HAVING Mnh HAVING thng c s dng cng mnh group by. Sau HAVING l biu thc iu kin. Biu thc iu kin ny khng tc ng vo tng bn ghi ca ton bng c ch ra trong mnh from m ch tc ng ln lt tng nhm cc bn ghi ch ra ti mnh group by. V d 4.5: Tm s hiu nhng nh cung cp cung cp hn hai mt hng. SELECT SNo FROM SP GROUPE BY SNo HAVING COUNT (DISTINCT PNo) >2 4.6. Truy vn thng tin t nhiu bng d liu(php kt ni). Trong php ni, cc ct tham gia kt ni phi c min tr l snh c vi nhau. Tn ct ca cc bng khc nhau c th vit tng minh qua tn bng (Tn_bng.Tn_ct). V d 4.6: Vi mi mt hng c cung cp, cho bit s hiu ca mt hng, s hiu, tn v a ch ca hng cung cp mt hng . SELECT PNo, SNO, SNAME, CITY FROM SP, S WHERE SP.SNo = S. SNo Ch : - Trong cc php tm kim c hn mt bng, nu tn cc ct l khng duy nht th bt buc phi vit tn ct dng tng minh. Trng hp khng c mnh WHERE khi php tch - cc s c thc hin. - iu kin trong mnh WHERE thng l iu kin kt ni d liu hoc l iu kin tm kim d liu. 4.7. Truy vn lng nhau V d 4.7.1: Tm tn nhng nh cung cp cung cp mt hng c s hiu P2. SELECT DISTINCT SNAME FROM S WHERE SNO IN (SELECT SNo FROM SP WHERE PNo = P2) V d 4.7.2: Tm tn cc nh cung cp khng cung ng mt hng c s hiu P1. SELECT DISTINCT SNAME FROM S WHERE SNO NOT IN (SELECT SNo FROM SP WHERE PNO=P1) Ch : V c tnh s dng cu hi con. - Php lng nhau c th c lng nhiu mc - Cu hi con phi t trong hai du ngoc n trong mnh WHERE hoc HAVING v i sau ton t so snh d liu. - Ni dung ch nh trong mnh SELECT ca cu hi con ch l mt tn (Mt ct hoc mt biu thc) - Kt qu ca cu hi con phi c kiu tng thch vi query m. Trong Query con khng c s dng cc php ton so snh Like, Between. Kt qa chn lc d liu trong Query con ch dng cho vic so snh ca Query m m khng c a ra bng kt qu v vy trong Query con khng c s dng mnh ORDER BY. - Ngi ta thng dng Query con khi gi tr so snh cho vic tm kim trong Query m l cha tng minh. V d 4.7.3: Cho bit s hiu nh cung cp, s hiu mt hng ca cc mt hng c cung cp vi s lng ln nht. SELECT DISTINCT SNo, PNo FROM SP WHERE QTY = (SELECT MAX(QTY ) FROM SP) V d 4.7.4: Cho bit s hiu nh cung cp, tn nh cung cp cha cung cp mt hng no. SELECT SNo,SNAME FROM S WHERE SNo NOT IN (SELECT SNo FROM SP ) - Tm kim vi t kho [NOT] EXISTS V d 4.7.5: a ra tt c cc thng tin v cc nh cung cp cung cp t nht mt mt hng no . SELECT * FROM S WHERE EXISTS ( SELECT SNo FROM SP,S WHERE SP.SNo = S. SNo ) V d 4.7.6: a ra tt c cc thng tin v cc nh cung cp cha cung cp bt c mt mt hng no. SELECT * FROM S WHERE NOT EXISTS ( SELECT SNo FROM SP,S WHERE SP.SNo = S. SNo ) - Tm kim c s dng lng t ANY v ALL. V d 4.7.7: a ra tn mt hng, mu, trng lng ca nhng mt hng c cung cp bi nh cung cp c s hiu S1. SELECT PNAME, COLOR, WEIGHT FROM P WHERE PNo = ANY( SELECT PNo FROM SP WHERE SNo = S1) V d 4.7.8: Tm s hiu nh cung cp cung cp mt mt hng no vi s lng ln hn hoc bng s lng mi ln cung cp mt mt hng ca cc nh cung cp. SELECT SNo FROM SP WHERE QTY >= ALL (SELECT QTY FROM SP) Mnh trn hon ton tng ng vi SELECT SNO FROM SP WHERE QTY = (SELECT MAX (QTY) FROM SP ) Ch : Trong nhiu trng hp c th thay th gia cc mnh exits vi in hoc not in, equal any tng ng vi not equal all. v.v... V d 4.7.9: V d 3.22 c th tng ng vi : SELECT * FROM S WHERE SNo NOT IN (SELECT SNo FROM SP) Hoc SELECT * FROM S WHERE SNo <> ALL (SELECT SNo FROM SP) - Tm kim c cha php tnh tp hp V d 4.7.10: Tm s hiu nhng nh cung cp hin thi cha cung cp mt mt hng no c. SELECT SNo FROM S MINUS (SELECT SNo FROM SP) V d 4.7.11: Tm s hiu cc mt hng c cung cp bi cc nh cung cp c s hiu l S1 hoc S2. (SELECT PNo FROM SP WHERE SNo=S1) UNION (SELECT PNo FROM SP WHERE SNo=S2) 4.8.Cc hm tnh ton trn nhm cc bn ghi Trong SQL c cc hm mu gm: Count (tn_ct): m s hng trong bng d liu ngun (khng m nhng hng c gi tr NULL trong ct ch nh). c bit count(*): m s hng tho yu cu tm kim m khng cn quan tm ti bt k mt ct no. Max(tn_ct | biu_thc_s) : Tm gi tr ln nht ca ct kiu s hoc biu thc s. Min(tn_ct | biu_thc_s): Tm gi tr nh nht ca ct kiu s hoc biu thc s. Sum(tn_ct | biu_thc_s): Tnh tng ca ct kiu s ch nh hoc biu thc s. Avg(tn_ct | biu_thc_s): Tnh gi tr trung bnh ca ct kiu s hoc biu thc s. V d 4.8.1: Cho bit s ln mt hng c s hiu P2 c cung cp. SELECT COUNT ( * ) FROM SP WHERE PNO = P2 V d 4.8.2: Tm hiu s s lng mt hng c s hiu mt hng l P1 cung cp mt ln nhiu nht v mt ln t nht ca nh cung cp c s hiu l S1. SELECT MAX (QTY) MIN (QTY) FROM SP WHERE SNO = S1 AND PNO = P1 V d 4.8.3: Tm tng s tin, gi trung bnh cc mt hng c cung cp bi nh cung cp c s hiu l S1. SELECT SUM(QTY*PRICE), AVG(FRICE) FROM SP WHERE SNO = S1 V d 4.8.4: Cho bit s lt cc mt hng khc nhau c cung cp. SELECT COUNT (DISTINCT PNO ) FROM SP Ch : Biu_thc_iu_kin trong mnh WHERE l mt biu thc Logic v biu thc ny tc ng ln tng bn ghi ca ton bng ch ra trong mnh FROM. 4.9. Cc hm tnh ton trn bn ghi 4.9.1.Cc hm ton hc ABS(x): Tr tuyt i ca x SQRT(x): Cn bc hai ca x LOG(x): Logarit t nhin ca x EXP(x): Hm m c s e ca x SIGN(x): Ly du ca s x (-1: x<0, 0:x=0, +1: x>0) ROUND(x,n): Lm trn ti n s l 4.9.2.Cc hm x l chui k t LEN(str): Cho chiu di dy k t str LEFT(str,n): Ly n k t pha tri ca dy str RIGHT(str,n): Ly n k t pha phi ca dy str MID(str,p,n): Ly n k t ca dy str k t v tr p trong dy. 4.9.3.Cc hm x l ngy thng v thi gian DATE(): Cho ngy thng nm hin ti DAY(dd): Cho th t ngy trong thngca biu thc ngy dd MONTH(dd): Cho th t thng trong thngca biu thc ngy dd YEAR(dd): Cho nm ca biu thc ngy dd HOUR(tt): Cho gi trong ngy MINUTE(tt): Cho s pht ca thi gian tt SECONDS(tt): Cho s giy ca biu thc gi tt Cu hi v bi tp chng 3 3.1 Nu cc cu lnh nh ngha d liu trong ngn ng SQL. 3.2 Nu cc cu lnh thao tc d liu trong ngn ng SQL. 3.3 Nu cc cu lnh an ton d liu trong ngn ng SQL. 3.4 Cho CSDL c 3 bng: DS( Sbd, Hoten, Ngaysinh, Gioitinh, Quequan) SBD_PH( Sbd,Sophach) DTM( Sophach, Diem) Trong : Sbd: s bo danh Hoten: H v tn sinh vin Ngaysinh: Ngy sinh Gioitinh: Gii tnh Quequan: Qu qun Sophach: S phch Diem: im thi ca mn hc Hy dng cc cu lnh ca SQL thc hin cc cng vic sau: Nu cc cu lnh to 3 bng trn v chn kiu d liu thch hp cho cc ct. S dng cu lnh vo d liu, nhp vo 3 bng trn mt s bn ghi. Xo sinh vin c c s bo danh 123. Cho bit h v tn, ngy sinh, gii tnh, qu qun, im thi ca mi sinh vin. Cho bit h v tn, ngy sinh, gii tnh, qu qun ca cc sinh vin c h Trn. Cho bit h v tn, ngy sinh, gii tnh, qu qun ca cc sinh vin c tn l Hng. Cho bit h v tn, ngy sinh, gii tnh, im thi ca nhng sinh vin c im thi ( 5. Cho bit h v tn, ngy sinh, im thi ca nhng sinh vin c gii tnh l Nu v qu Thai Binh. Danh sch a ra sp xp theo chiu gim dn ca im. Cho bit h v tn, ngy sinh, im thi ca nhng sinh vin c qu Nam Dinh hoc Thai Binh v c im thi < 3 hoc im thi >8. Danh sch a ra sp xp theo chiu gim dn ca im. Cho bit h v tn, ngy sinh, gii tnh, qu qun ca cc sinh vin khng i thi. a ra s lng sinh vin, im trung bnh ca cc sinh vin mi tnh. a ra tng s lng sinh vin n, im trung bnh ca cc sinh vin n mi tnh. Danh sch a ra sp xp theo gim dn ca im trung bnh. a ra tng s, im trung bnh ca cc sinh vin n mi tnh c im trung bnh >7. Danh sch a ra sp xp theo gim dn ca s lng sinh sinh vin, nu s lng bng nhau th sp xp theo chiu tng dn ca im trung bnh. Tng im thi thm 2 im cho cc hc sinh qu Sn La. 3.5 Cho CSDL c 3 bng: MatHang(Mamh, Tenmh, Mau, DVT) KhHang(Makh, Tenkh, Diachi, DT, Gioitinh) MuaBan(Mamh, Makh, Muaban, NgayMB, Soluong, Dongia) Trong : - Mamh: M mt hng Tenmh: Tn mt hng Mau: Mu mt hng DVT: n v tnh Makh: M khch hng Tenkh: Tn khch hng Diachi: a ch khch hng DT: in thoi Gioitinh: Gii tnh MuaBan: Mua bn trong khch hng mua th ghi l. F., khch hng bn th ghi l. T. NgayMB: Ngy mua hoc bn Soluong: S lng Dongia: n gi Hy dng cc cu lnh ca SQL thc hin cc cng vic sau: 1) Nu cc cu lnh to 3 bng trn v chn kiu d liu thch hp cho cc ct. S dng cu lnh vo d liu, nhp vo 3 bng trn mi bng mt s bn ghi. Cho bit makh ca cc khch hng bn hng. Cho bit makh, tenkh, tenmh ca nhng khch hng bn mt hng c mamh = MH001 hoc mamh = MH002. Cho bit thng tin ca cc mt hng m khch hng mua c n gi nh hn n gi trung bnh mi ln mua mt mt hng. Cho bit makh ca nhng khch hng bn mt hng mu vi s lng >100 trong qu I nm 2003. Cho bit tenkh, diachi, DT ca nhng khch hng bn mt hng mu Vang hoc xanh vi s lng >100. Cho bit tenkh, tenmh ca nhng khch hng c gii tnh l nam bn mt hng mu Den v mua mt hng mu Xanh vi 200> soluong>100. Cho bit cc thng tin ca cc khch cha tham gia mua bn ln no. Cho bit cc thng tin ca cc khch cha mua hng ln no. Cho bit cc thng tin ca cc mt hng cha c mua bn ln no. 12) Cho bit cc thng tin ca cc khch, s ln mua v s tin mua hng c s ln mua ln nht hoc nh nht trong thng 1 nm 1999. 13) Cho bit cc thng tin ca cc khch mua hng c s ln mua ln hn 3. Danh sch a ra sp xp theo chiu gim dn ca s ln mua. Cho bit mamh, tenmh, s lng tn kho ca mi mt hng. Xo i mt hng c mamh l 001. Thm ct thanhtien v tnh s tin mi ln mua bn mt mt hng. Cho bit thng tin v nhng khch hng bn mt hng mu xanh nhng khng mua mt hng mu . Chng IV: Rng buc ton vn v ph thuc hm 1.Cc vn lin quan n rng buc ton vn 1.1. nh ngha Trong mi CSDL lun tn ti nhiu mi lin h gia cc thuc tnh, gia cc b; s lin h ny c th xy ra trong cng mt quan h hoc trong cc quan h ca mt lc CSDL. Cc mi lin h ny l nhng iu kin bt bin m tt c cc b ca nhng quan h c lin quan trong CSDL u phi tho mn mi thi im. Nhng iu kin bt bin c gi l rng buc ton vn.. Trong thc t rng buc ton vn l cc quy tc qun l c p t trn cc i tng ca th gii thc. Chng hn mi sinh vin phi c mt m sinh vin duy nht, hai th sinh d thi vo mt trng phi c s bo danh khc nhau, mt sinh vin d thi mt mn hc khng qu 3 ln, Nhim v ca ngi phn tch thit k l phi pht hin cng y cc rng buc ton vn cng tt v m t chng mt cch chnh xc trong h s phn tch thit k - l mt vic lm rt quan trng. Rng buc ton vn c xem nh l mt cng c din t ng ngha ca CSDL. Mt CSDL c thit k cng knh nhng n th hin c y ng ngha ca thc t vn c gi tr cao hn rt nhiu so vi mt cch thit k gn nh nhng ngho nn v ng ngha v thiu cc rng buc ton vn ca CSDL. Cng vic kim tra rng buc ton vn thng c tin hnh vo thi im cp nht d liu ( thm, sa, xo). Nhng rng buc ton vn pht sinh phi cn c ghi nhn v x l mt cch tng minh (thng l bi mt hm chun hoc mt on chng trnh). Rng buc ton vn v kim tra s vi phm rng buc ton vn l hai trong s nhng vn quan trng trong qu trnh phn tch thit k c s d liu, nu khng quan tm ng mc n nhng vn trn, th c th dn n nhng hu qu nghim trng v tnh an ton v ton vn d liu, c bit l i vi nhng c s d liu ln. 1.2.iu kin iu kin ca rng buc ton vn l s m t, v biu din hnh thc ni dung ca n iu kin ca mt rng buc ton vn R c th c biu din bng ngn ng t nhin, ngn ng i s quan h, ngn ng m gi, ngn ng truy vn SQL, ngoi ra iu kin ca rng buc ton vn cng c th c biu din bng ph thuc hm (khi nim ph thuc hm s c cp trong chng 5) Sau y l mt s rng buc ton vn trn lc CSDL qun l sinh vin . Mi lp hc phi c mt m s duy nht phn bit vi cc lp hc khc trong trng. Mi lp hc phi thuc v mt khoa ca trng. Mi sinh vin c mt m s sinh vin duy nht, khng trng vi bt c sinh vin no trong trng. Mi hc vin phi ng k vo mt lp hc trong trng. Mi hc vin ch c thi ti a 3 ln cho mi mn hc. Tng s hc vin ca mt lp phi ln hn hoc bng s lng m c ca mt lp ti mt thi im no . 1.3..Bi cnh Bi cnh ca rng buc ton vn l nhng quan h m rng buc c hiu lc hay ni mt cch khc, l nhng quan h cn phi c kim tra khi tin hnh cp nht d liu. Bi cnh ca mt rng buc ton vn c th l mt hoc nhiu quan h. Chng hn vi rng buc ton vn R trn th bi cnh ca n l quan h Sinhvien 1.4.Bng tm nh hng Trong qu trnh phn tch thit k mt CSDL, ngi phn tch cn lp bng tm nh hng cho mt rng buc ton vn nhm xc nh thi im cn phi tin hnh kim tra khi tin hnh cp nht d liu. Thi im cn phi kim tra rng buc ton vn chnh l thi im cp nht d liu. Mt bng tm nh hng ca mt rng buc ton vn c dng sau: Tn RBTVThm(T)Sa(S)Xo(X)r1+r2-r3-(*)rnBng ny cha ton cc k hiu + , hoc -(*). Chng hn + ti (dng r1, ct Thm) th c ngha l khi thm mt b vo quan h r1 th RBTV b vi phm. Du Ti (dng r2, ct sa) th c ngha l khi sa mt b trn quan h r2 th RBTV khng b vi phm. , Quy c: -Khng c sa thuc tnh kho. -Nu khng b vi phm do khng c php sa i th k hiu l -(*). 1.5.Hnh ng cn phi c khi pht hin c RBTV b vi phm: Khi mt rng buc ton vn b vi phm, cn c nhng hnh ng thch hp. Thng thng c 2 gii php: Th nht: a ra thng bo v yu cu sa cha d liu ca cc thuc tnh cho ph hp vi quy tc m bo tnh nht qun d liu. Thng bo phi y v phi thn thin vi ngi s dng. Gii php ny l ph hp cho vic x l thi gian thc. Th hai:T chi thao tc cp nht. Gii php ny l ph hp i vi vic x l theo l. Vic t chi cng phi c lu li bng nhng thng bo y , r rng v sao thao tc b t chi v cn phi sa li nhng d liu no ? Kha ni, kho ngoi, gi tr NOT NULL l nhng rng buc ton vn min gi tr ca cc thuc tnh. Nhng rng buc ton vn ny l nhng rng buc ton vn n gin trong CSDL. Cc h qun tr c s d liu thng c cc c ch t ng kim tra cc rng buc ton vn v min gi tr ca kho ni, kho ngoi, gi tr NOT NULL. Vic kim tra rng buc ton vn c th tin hnh vo nhng thi im sau y. Th nht: Kim tra ngay sau khi thc hin mt thao tc cp nht CSDL. Thao tc cp nht ch c xem l hp l nu nh n khng vi phm bt c mt rng buc ton vn no, ngha l n khng lm mt tnh ton vn ca CSDL. Nu vi phm rng buc ton vn , thao tc cp nht b coi l khng hp l v s b h thng hu b (hoc c mt x l thch hp no ) Th hai: Kim tra nh k hay t xut, ngha l vic kim tra rng buc ton vn c tin hnh c lp vi thao tc cp nht d liu. i vi nhng trng hp vi phm rng buc ton vn, h thng c nhng x l ngm nh hoc yu cu ngi s dng x l nhng sai st mt cch tng minh. 2. Cc loi rng buc ton vn Trong qu trnh phn tch thit k CSDL, ngi phn tch phi pht hin tt c cc rng buc ton vn tim n trong CSDL . Vic phn loi cc rng buc ton vn l rt c ch, n nhm gip cho ngi phn tch c c mt nh hng pht hin cc rng buc ton vn, trnh b st. Cc rng buc ton vn c th c chia lm cc loi chnh nh sau: 2.1. Rng buc ton vn v min gi tr Rng buc ton vn c lin quan n min gi tr ca cc thuc tnh trong mt quan h. Rng buc ny thng gp. Mt s h qun tr CSDL t ng kim tra mt s rng buc loi ny. V d: Vi r l mt quan h ca Hoadon ta c rng buc ton vn sau:  2.2. Rng buc ton vn lin thuc tnh Rng buc ton vn lin thuc tnh l mi lin h gia cc thuc tnh trong mt lc quan h. V d: Vi r l mt quan h ca Hoadon ta c rng buc ton vn sau:  2.3. Rng buc ton vn lin b lin thuc tnh Rng buc loi ny l mi lin h gia cc thuc tnh trong nhiu lc quan h. V d: Vi r, s ln lt l quan h ca Dathang, Hoadon ta c rng buc ton vn sau:  2.4. Rng buc ton vn v ph thuc tn ti Rng buc ton vn v ph thuc tn ti cng c gi l rng buc ton vn v kho ngoi. Cng ging nh rng buc ton vn v kho chnh, rng buc ny rt ph bin trong CSDL. V d: Vi r, s ln lt l mt quan h ca Dathang, Khach ta c rng buc ton vn sau:  2.5. Rng buc ton vn tng hp (lin b - lin quan h) Rng buc ny c xc nh trong trng hp mi thuc tnh A ca mt lc quan h Q c tnh ton gi tr t cc thuc tnh ca cc lc quan h khc. 3.ph thuc hm Quy c v k hiu Cc ch hoa u b cc ch ci in hoa nh A, B, C,. .. biu th mt thuc tnh n. Cc ch hoa cui b cc ch ci in hoa nh U, V,. .., X, Y, Z biu th mt tp thuc tnh, c th l tp ch c mt thuc tnh. R c dng biu th mt lc quan h. Cng c th t tn cc quan h bng lc ca chng. Chng hn mt quan h c cc thuc A,B, C c th c vit l ABC hoc R(ABC ). S dng r cho mt quan h, l th hin hin hnh ca lc quan h R. K hiu ni kt chui c dng biu th cho php hp. Do A1A2. ..An c dng biu din tp cc thuc tnh { A1, A2,. .., An } v XY vit tt ca X(Y. Trng hp XA hay AX cng c vit thay cho X({A} hay {A}(X, vi X l tp cc thuc tnh v A l mt thuc tnh n. 3.1.nh ngha v biu din ph thuc hm nh ngha 4.1 Cho R l mt lc quan h vi U = {A1, A2,..., An} l tp thuc tnh. X v Y l tp con ca U. Ni rng X ( Y ( c l X xc nh hm Y hoc Y ph thuc hm vo X ) nu vi bt k quan h r no l gi tr hin hnh ca R v bt k hai b t1, t2 ( r m t1[X] = t2[X] th t1 [Y]= t2[Y] Ph thuc hm k hiu l FD, cn ch rng ch xt cc ph thuc hm tho mn cho mi quan h trn lc quan h tng ng ca n. Khng th xem xt mt ph thuc hm tho mt quan h r c bit (v d quan h rng) ca lc R ri sau qui np rng ph thuc hm l tho trn R. V d 4.1: (Trong lc CSDL ST) * Mt s ph thuc hm c bn nht l nhng ph thuc hm khng nh rng mt kho xc nh c tt c cc thuc tnh ca lc quan h Chng hn trong SUPPLIERS c SNO ( SNAME, SNO ( SADDR. Hay trong SUPPLIES c SNO INO ( PRICE Cn trong CUSTOMERS c CNO ( CNAME CADDR BALANCE * Mt s ph thuc hm tm thng nh SNAME ( SNAME SADDR ( SADDR SNAME SADDR ( SADDR * Mt s ph thuc hm khc SNO INO ( SADDR SNAME INAME 3.2.Bao ng ca tp ph thuc hm v h lut dn Armstrong 3.2.1.Bao ng ca tp ph thuc hm Gi F l tp tt c cc ph thuc hm i vi lc quan h R v X ( Y l mt ph thuc hm vi X,Y ( U. Ni rng X ( Y c suy din logic t F nu mi quan h r tho cc ph thuc hm ca F th cng tho X ( Y. Chng hn F=(A ( B, B (C( th A ( C c suy ra t F. Gi F+ l bao ng (closure) ca F, tc l tt c cc ph thuc hm c suy din logic t F. Nu F = F+ th F l h y (full family) ca cc ph thuc hm. 3.2.2.H lut dn Armstrong c th xc nh kho ca mt lc quan h v hiu c cc php suy din logic cho cc ph thuc hm cn tnh c F+ t F, hoc t nht phi khng nh c X ( Y c thuc F hay khng nu bit ph thuc hm X ( Y v tp ph thuc hm F. Do i hi phi c nhng qui tc suy din cho bit lm sao c th suy ra mt hay nhiu ph thuc hm t cc ph thuc hm khc. Tp cc qui tc ny c c Armstrong a ra nm 1974 v c gi l h tin Armstrong. Cho R l lc quan h vi U= (A1,..,An( l tp cc thuc tnh ca n v X, Y, Z, W(U. A1 (phn x): Nu Y ( X th X ( Y Quy tc ny a ra nhng ph thuc hm tm thng l nhng ph thuc hm m v tri cha v phi. S dng quy tc ny ch ph thuc vo U, khng phi vo F. A2 (tng trng) : Nu Z( U v X (Y th XZ ( YZ. Trong k hiu XZ thay cho k hiu X ( Z v X ( Y c th thuc F hoc c suy din logic t F. A3 (bc cu): nu X ( Y v Y ( Z th X ( Z. V d 4.2: Cho lc quan h R(ABCD) vi cc ph thuc hm A ( C v B ( D Chng minh rng AB ( ABCD Tht vy ta c A ( C (gi thit) ( AB ( ABC (lut tng trng) v B ( D (gi thit) (ABC ( ABCD (lut tng trng) Vy AB ( ABCD (lut bc cu) Tnh ng n ca h tin Armstrong B 4.1 H tin Amrstrong l ng. C ngha l F l tp cc ph thuc hm ng trn quan h R. Nu X(Y l mt ph thuc hm c suy dn t F nh h tin Amrstrong th X( Y l ng trn quan h R. Chng minh: Ln lt kim tra tnh ng n ca ba b A1, A2, A3 A1: Tin A1 r rng l ng v khng th c hai b bng nhau trn X m li khng bng nhau trn tp con ca n. A2: Gi s rng quan h r tho X(Y v tn ti hai b t, u ( r sao cho t[XZ] = u[XZ} m t[YZ} ( u[YZ}. V rng t[Z] = u[Z] nn t[YZ} ( u[YZ th t[Y] ( u[Y]. Nhng v t[X]=u{X] nn t{Y] ( u{Y] l tri vi gi thit X(Y.Vy t[YZ}=u{YZ}. A3: Cho X(Y v Y(Z ng trn quan h r. Gi s tn ti hai b t v u(r sao cho t{X} = u{X} v t[Z}( u[Z]. T X(Z suy ra v t{X} = u[X] nn t[Y] = u[Y] Nhng li c t[Y] = u[Y] v t[Z] ( u[Z] l tri vi gi thit l Y(Z. Vy t[Z] = u[Z]. Suy ra X(Z ng trn quan h r. T h tin Armstrong suy ra mt s lut sau y: Cc quy tc suy din b sung B 4.2 a. Lut hp: Nu X(Y v X(Z th X(YZ b. Lut ta bc cu: Nu X(Y v WY(Z th XW(Z c. Lut tch: Nu X(Y v Z (Y th X(Z Chng minh: a. T X(Y dng lut tng trng, thm X c X(XY T X(Z dng lut tng trng thm Y c XY(Yz v cui cng dng lut bc cu suy ra v X(Xy v XY(XZ nn X(YZ. b. T X(Y, dng lut tng trng, thm W c WX (WY. Dng lut bc cu cho WX ( WY v WY(Z suy ra WX(Z. c. V Z (Y nn Y(Z (theo lut phn x ). Dng lut bc cu cho X(Y v Y(Z c X(Z. Mt h qu quan trng ca lut tch v lut hp l nu X( Y suy ra X(Ai vi mi Ai(Y. 3.3.Bao ng ca tp thuc tnh d dng cho vic chng minh tnh y ca h tin Arsmtrong. y a thm khi nim bao ng (closure) ca tp cc thuc tnh i vi tp cc ph thuc hm. Gi F l tp cc ph thuc hm trn tp thuc tnh U, X(U. X+ l bao ng ca X (i vi F) c nh ngha nh sau: X+= {A( X(A(F+} Ni c th : X+ l tp tt c cc thuc tnh A m ph thuc hm X(A c th c suy din logic t F nh h tin Armstrong. B 3.3 X(Y suy dn t h tin Armstrong khi v ch khi Y( X+. Chng minh : o: Gi s Y=A1... An vi A1,. .., An l cc thuc tnh v Y( X+ suy ra Ai ( X+ T nh ngha X+ c X( Ai. p dng tin Armstrong cho mi Ai( Y, lut hp suy ra X(Y Thun: Gi s c X(Y, p dng h tin Armstrong cho mi i c X(Ai, Ai(Y nh lut tch suy ra Ai (X+. T suy ra Y( X+. Suy ra : chng minh X(Y, ngoi cch dng nh ngha ph thuc hm, p dng h tin Armstrong v cc quy tc b sung cn c th p dng b trn bng cch: -Tnh X+. - Khng nh Y( X+. nh l 3.3 H tin Armstrong l ng v y . Chng minh : Tnh ng n ca h tin c chng minh qua b 4.1. y ch cn chng minh tnh y tc l nu X( Y khng tho trn r th X(Y khng th suy dn logic t F. Gi F l tp cc ph thuc hm trn tp thuc tnh U. Gi s rng X(Y l khng th suy dn c t h tin Armstrong. Xt quan h r gm hai b phn nh sau: 11...1 11...1 11...1 00...0 Cc thuc tnh thuc X Cc thuc tnh cn li Trc ht cn ch ra rng tt c cc ph thuc hm thuc F u tho r. Tht vy, gi s (V ( W) ( F nhng khng tho trn r. Do V ( X+, hoc hai b ca r s khng bng nhau t nht trn mt thuc tnh ca V nh vy W khng th l tp con ca X+ hoc V ( W tho trn r. Gi A ( W nhng A khng thuc X+. V XV ( X+, X( V suy ra t b 4.3. ( V(W ) ( F do vy, nh lut bc cu suy ra X( A. Nhng do A khng thuc X+ nh gi thit do vy dn n mu thun. T n kt lun rng mi (V(W) ( F u tho trn r. By gi cn chng minh rng X(Y khng tho trn r. Gi s rng X(Y l tho trn r. Nh trn c X ( X+ v suy ra Y ( X+, nu khng hai b thuc r l bng nhau trn X nhng khng bng nhau trn Y. Theo b 4.3 th X(Y c th suy ra t h tin , i l hon ton mu thun. Do vy X(Y khng th ng trn r. n y c th kt lun: Nu X(Y khng suy dn c t tin Arsmtrong th X(Y khng th suy dn c t F. Vy h tin l y . Tnh ton bao ng Vic tnh bao ng F+ ca tp cc ph thuc hm F trong trng hp tng qut l rt kh khn v tn km thi gian bi v tp cc ph thuc hm F+ rt ln cho d F c th l nh. Chng hn F = { A( B1, A( B2,. .,A( Bn }. F+ khi cn c tnh c nhng ph thuc hm A(Y vi Y( {B1, , Bn}. Nh vy s c 2n tp con ca Y nn c 2n ph thuc hm. Nhng tnh X+, bao ng ca tp thuc tnh X li khng kh. Theo b 4.3 c th kim tra c X(Y ( F+ hay khng bng cch tnh X+ ng vi F. Vic tnh bao ng X+ c th hin qua thut ton sau y: Thut ton 4.1: Tnh bao ng ca tp cc thuc tnh i vi mt tp cc ph thuc hm. Vo: tp U hu hn cc thuc tnh, tp cc ph thuc hm F trn U v X ( U. Ra: X+, bao ng ca X i vi F. Phng php: Tnh lin tip tp cc thuc tnh X0, X1, . .. theo quy tc: 1) X0 = X 2) Xi+1= Xi ( A sao cho ( (Y( Z) (F, A(Z v Y (Xi Bi v X = X0 (. .. ( U, U l hu hn nn s tn ti mt ch s i m Xi = Xi+1. Khi X+ = Xi. V d 4. 3: cho F l tp tm ph thuc hm sau: AB( C D( EG ACD( B C ( A BE ( C CE (AG BC ( D CG (BD V X = BD. Tnh X+ p dng thu t ton 4.1: t X0 = BD. Mun tnh X1. Hy chn cc ph thuc hm c v tri l con ca BD (v tri l B, D hoc BD) v kt np cc v phi ca chng vo X0. y ch c D ( EG. Khi c X1 = BDEG Tng t tm cc ph thuc hm c v tri l con X1 ( c D ( EG, BE ( C ). Khi c X2 = BCDEG v.v. Tip tc s c X3 = X4 = ABCDEG v cui cng X+ = (BD)+ = ABCDEG. V d 4.4: Cho lc quan h R v tp cc ph thuc hm F= {AB -> E, AG -> I, BE -> I, E -> G, GI -> H} Chng minh rng: AB -> GHI. Gii -Tnh AB+ t XO = AB Ta c X1 = ABE (v AB -> E) Ta c X2 = ABEIG (v AB -> E, BE -> I, E -> G) Ta c X3 = ABEIGH (v AB -> E, AG -> I, BE -> I, E -> G, GI -> H) Ta c X4 = ABEIGH (v AB -> E, AG -> I, BE -> I, E -> G, GI -> H) V X3= X4 = ABEIGH ( GHI nn AB -> GHI. Sau y cn phi ch ra rng thut ton 4.1 l ng. nh l 4.2 Thut ton 4.1 tnh X+ l ng. Chng minh: Chng minh bng quy np. Bc c s: ng v A ( X r rng X ( A. Bc quy np: Gi s bc j - 1 ng. Cn chng minh cho bc th j. Tc l, nu A thm vo Xj th A (X+. Xj -1 ch cha cc thuc tnh X+, A s l thuc tnh c a vo Xj. Tht vy A ( Z, (Y(Z) ( F v Y ( Xj -1, Y ( X+ theo gi thit quy np. X (Y v Y ( Z c X ( Z, Z ( A (theo lut phn x) v X ( A (theo lut bc cu) v do A ( X+. Ngc li, cn chng minh rng nu A ( X+ th A phi thuc vo Xi no . C iu khng quan trng l thut ton 4.1 c th kt thc sm hn trc khi tnh ton bc th j cho Xj. Nu thut ton dng bc Xi = Xj +1 vi i < j th r rng rng Xi = Xj. Do vy Xi = X+, trong c c thuc tnh A. Nu trong qu trnh chng minh cn s dng ti h tin Armstrong: X ( Y suy dn t F th mi thuc tnh A ( Y c thm vo ti mi Xj no . Cc bc quy np s thc hin thm mt s dng, trong mi dng l mt ph thuc hm thuc F v s dng lut phn x hoc gi thit ca bc quy np trc hoc s dng lut tng trng v lut bc cu. Cui cng s l X( Y. 3.4.Ph v tng ng 3.4.1.nh ngha: Cho F v G l hai tp ph thuc hm trn tp thuc tnh U, ta ni rng tp ph thuc hm F tng ng vi tp ph thuc hm G nu v ch nu F+ = G+ Nu F+ = G+ th ta ni rng F l ph ca G v ngc li F l ph ca F. Thut ton xc nh F v G c tng ng khng?  3.4.2.Ph ti thiu Gi tp cc ph thuc hm F l ti thiu nu: Mi v phi ca mt ph thuc hm thuc F ch c mt thuc tnh. Khng tn ti mt ph thuc hm X(A thuc hm F v mt tp con Z ca X m: F+ = (F - (X (A( ( (Z ( A()+ Khng tn ti mt ph thuc hm X(A thuc F m F+ = (F - {X (A} )+ iu kin b) bo m khng c mt thuc tnh no tham gia pha tri ca ph thuc hm l d tha v php kim tra cc ph thuc hm d tha v tri nh sau: Thuc tnh B trong X i vi ph thuc hm X(A l d tha tha nu v ch nu A thuc (X-{B})F+. V trc quan, iu kin c) bo m cho tp F khng c mt ph thuc hm no l d tha v kim tra X(A c d tha hay khng bng cch tnh X+ ng vi F - {X (A}. V phi ca ph thuc hm iu kin a) ch c mt thuc tnh bo m chc chn khng c mt thuc tnh no v phi l d tha. Nu G l mt tp ph hm ti thiu theo ngha trn v G tng ng F th ta ni rng G l ph ti thiu ca F. Ch : Nu ta thc hin b) trc ri mi thc hin c) th ta tm c ph ti thiu, thc hin ngc li th cha chc ng. nh l 3.4.2 Mi tp ph thuc hm F u tng ng vi mt tp F ti thiu. Chng minh Theo b 4.4, gi s rng khng v phi no ca cc ph thuc hm ca F c nhiu hn mt thuc tnh. kim tra iu kin b) xt cc ph thuc hm (X( Y) ( F. Nu ( F - (X(Y()+ = F+ th loi b X(Y khi F. kim tra iu kin c) xt cc ph thuc hm (X( A) ( F. Nu B( X v (X-B) ( A th loi b B khi X. Cui cng ta thu c F ti thiu. Ch rng cc ph thuc hm F c sp xp theo mt th t khc nhau th s cho kt qu khc nhau. V d 3.4.2.1 : Cho tp F gm: A ( B A( C B ( C B ( A C ( A C th loi b khi F: B ( A v A ( C hoc loi b B ( C nhng khng th ng thi loi b c ba ph thuc hm. Nh vy iu kin b) c tho. Cn kim tra iu kin c) cho tp cc ph thuc cn li ca F. y cng cn lu ti th t cc thuc tnh xut hin bn v tri ca cc ph thuc hm. Loi b cc thuc tnh v tri ca cc ph thuc hm sao cho tp cc thuc tnh vn l tng ng. Qu trnh tip tc cho n khi khng th loi b c thuc tnh no na. Nh vy tp cc ph thuc cn li ca F s to nn F v tho ba iu kin ca mt tp ti thiu. V d 3.4.2.2: Cho tp F: AB ( C D ( E CG ( D C ( A D ( G CE ( A BC ( D BE ( C CE ( G ACD ( B CG ( B p dng thut ton 4.1 cho tp F theo th t t phi qua tri v t di ln trn. R rng CE ( A l d tha v c th suy ra t C ( A. CG ( B l d tha v CG ( D, C( A v ACD ( B suy ra CG ( B. T d dng tnh c (CG)+ v khng cn ph thuc hm no l d tha. ACD ( B c th thay bi CD ( B v C( A. T c tp ti thiu l: AB ( C C ( A BC ( D CD ( B D ( E D ( G BE ( C CG ( D CE ( G Nu loi b CE ( A, CG ( D v ACD ( B s c mt tp ti thiu khc: AB(C D(G C(A BE(C BC(D CG ( B D ( E CE ( G Lu rng hai tp ti thiu nu trn c s lng cc ph thuc hm l khc nhau. 3.5.Thut ton xc nh kho ca lc quan h Cho lc quan h R(U). F l tp ph thuc hm ca R + K c gi l siu kho ca lc R nu K(U (K+=U) + K c gi l kho ca lc R nu: 1, K ( U (K+= U) 2, Vi mi K' ( K th K' ( U 3.5.1.Thut ton tm kho Vo: Lc quan h R vi tp thuc tnh U v tp ph thuc hm F. Ra: Tp K l kho ca R. Phng php: 1. t K = U 2. Lp li qu trnh loi khi K thuc tnh A m {K-A}+ = U. M t thut ton (ta Pascal) Begin K:= U; FOR each A IN K DO IF {K-A}+ = U THEN K := K-A END. Thut ton trn tm c mt kho ca R. Mun tm cc kho khc (nu c) ta c th thay i th t loi b cc thuc tnh trong K. V d 4. 5: Cho lc quan h R(ABCD) vi cc ph thuc hm A ( C v B ( D. Hy tm kho ca lc quan h trn. 1) K = U= abcd 2) Ln lt loi b cc thuc tnh trong K: Loi b D. Ta c abc+ = U (v A ( C v B ( D) nn K= ABC Loi b C. Ta c ab+ = U (v A ( C v B ( D) nn K= AB Loi b B. Ta c a+ = AC ( U (v A ( C ) nn K= AB Loi b A. Ta c B+ = BD ( U (v B ( D ) nn K= AB Vy K= AB. T thut ton tm kho ta c cc nhn xt sau: 1) Cc thuc tnh khng xut hin trong c v tri v v phi ca tp ph thuc hm F phi c trong kho. 2) Cc thuc tnh ch xut hin bn v tri ca cc ph thuc hm trong F cng phi thuc kho. 3) Trong qu trnh tm kho c th b i tt c cc thuc tnh n pha bn phi ca cc ph thuc hm trong F. Tuy nhin cn phi kim tra li v khng phi lc no cc thuc tnh cng b c Cc nhn xt ny gip c th pht hin ra kho mt cch nhanh chng. Chng hn cho lc quan h R vi tp thuc tnh{a, B, C, D} v tp ph thuc hm F = {A ( C, B ( D } c th pht hin ra ngay kho ca R l AB. 3.5.2. Thut ton tm tt c cc kho Thut ton c bn:   Thut ton ci tin: (Thut ton tm tt c cc kho ca mt lc quan h) Trc khi vo thut ton ci tin, ta cn quan tm mt s khi nim sau: + Tp thuc tnh ngun: (TN) cha tt c cc thuc tnh xut hin v tri v khng xut hin v phi ca cc ph thuc hm v cc thuc tnh khng xut hin c v tri ln v phi ca cc ph thuc hm. + Tp thuc tnh ch (TD): cha tt c cc thuc tnh c xut hin v phi v khng xut hin v tri ca cc ph thuc hm. + Tp thuc tnh trung gian (TG): cha tt c cc thuc tnh xut hin c v tri ln v phi ca cc ph thuc hm. H qu: Nu K l kho ca Q th TN ( K v TD( K =( Thut ton:  V d: (Theo v d trn)   Bi tp chng IV I. Bi tp ph thuc hm: Bi 4.1: Cho lc quan h ( = (U, F) vi: U = ABCDEGH v tp ph thuc hm F = {AB( C, B (D, CD( E, CE ( GH, G (A} f = AB( E, chng minh rng vi mi quan h R trn U nu R tho F th R cng tho f. Bi 4.2: Cho lc quan h ( = (U, F) vi: U = ABCDEGH v tp ph thuc hm F = {AB( E, AG (J, BE( I, E ( G, GI ( H} f = AB( GH, chng minh rng f suy dn c t F. Bi 4.3: Cho lc quan h ( = (U, F) vi: U = ABCDEG v tp ph thuc hm F = {A( C, BC (D, D( E, E ( A} Hy tnh: (AB)+ ; ((DE)+A)+ Bi 4.4: Cho lc quan h ( = (U, F) vi: U = ABCDEGH v tp ph thuc hm. F = {AB( GH, GD ( AHE, C( AGH, HE ( BC} Hy tnh: (CE)+ ; (CD)+ II. Bi tp v kho: Bi 4.5: Cho cc mnh sau, hy cho bit mnh no ng, mnh no sai. M1: K ( U l mt kho khi v ch khi K ( U. M2: K ( U l mt kho th K ( U. M3: Hai kho bt k l khng giao nhau. M4: Hai kho bt k l khng bao nhau. M5: Mi lc quan h u c t nht mt kho. M6: Bn thn U cng c th l mt kho. M7: Tn ti mt lc quan h khng c kho no. M8: U khng th l kho ca lc . M9: Hp ca hai kho phn bit l mt kho. M10: Hp ca hai kho l mt siu kho. Bi 4.6: Xy dng lc quan h c cc kho l ADE, BCH, CG, AHE. Bi 4.7: Cho lc quan h ( = (U, F) vi: U = ABCDEGHK v tp ph thuc hm F = {ADH( BC, GH (BE, D( CG, CH ( K} Hy tm: - Giao ca tt c cc kho. - Lc cho c mt hay nhiu kho. - Hy tm tt c cc kho ca lc . - Mt s cc phn t khng kho. Bi 4.8: Cho lc quan h ( = (U, F) vi: U = ABCDE v tp ph thuc hm F = {AB( C, AD (B, B( D} Hy tm cc phn t kho ca lc trn. Bi 4.9: Cho lc quan h ( = (U, F) vi: U = ABCDEGH v tp ph thuc hm F = {ABC ( DE, BCD ( G, ABH( EG, CE ( GH} Hy tm mt kho ca lc . Bi 4.10: Cho lc quan h ( = (U, F) vi: U = ABCD v tp ph thuc hm F = {AD ( BC, B ( A, D ( C} Hy tm cc kho ca lc . Cho bit C c phi l thuc tnh kho hay khng? Bi 4.11: Cho lc quan h ( = (U, F) vi: U = ABCDE v tp ph thuc hm F = {DE ( A, B ( C, E ( AD} Hy tm mt kho ca lc . Tp BCE c phi l kho ca ( khng? V sao? Tp AD c phi l kho ca ( khng? V sao? Tp BD c phi l kho ca ( khng? V sao? Bi 4.12: Cho lc quan h ( = (U, F) vi: U = ABCDEG v tp ph thuc hm F = {A ( BD, BC ( DE, D ( B, CD ( GE, BE ( A, G ( B} Hi rng tp CGE c l kho ca lc hay khng? Lc c mt hay nhiu kho? Hy tm mt kho v gii thch. Chng v: dng chun v cc vn chun ho lc c s d liu quan h 1.Dng chun 1.1.Thit k km gy nguy him cho CSDL Mt trong hai nguyn nhn sau y do thit k km s gy nguy him cho CSDL trng lp thng tin Khng c kh nng trnh by thng tin mt cch chc chn V d: Cho mt lc quan h dng ghi nhn gio vin v lp ging dy ca gio vin GIANGDAY(MONHOC, SOTIET, LOP, GV, HV, DC) Cc ph thuc hm: MONHOC (SOTIET.MONHOC, LOP( GV, GV(HOCVI.DC Xt mt tnh trng d liu nh sau: MONHOCSOTIETLOPGVHOCVIDCCSDL90TH1N.V.ACNQ1CTDL75TH1N.V.BTSQ2CSDL90TH2N.V.ACNQ1CTDL75TH2N.V.BTSQ2 CN: C nhn, TS: Tin s Do c ph thuc hm MONHOC (SOTIET nn s tit ca dng th 2 v dng th 3 gy trng lp thng tin. Tng t ph thuc hm GV(HOCVI.DC nn hc v v a ch ca dng th 2 v th 3 gy nn s trng lp thng tin.Cc d liu gy trng lp thng tin l cc d liu c th suy on c mt cch chc chn v duy nht t ph thuc hm. y lu hc v v a ch ca mt gio vin th gio vin phi tham gia ging dy mt lp no . gii quyt vn lu thng tin cc gio vin khng tham gia ging dy ngi ta dng gi tr NULL cho cc thuc tnh MONHOC, SOTIET, LOP. Nh vy, lc quan h nylu tr hai thng tin ca hai i tng khc nhau mt l ging dy ca cc ca cc gio vin tham gia ging dy, hai l thng tin ca cc gio vin khng tham gia ging dy. Vn ny sinh y l khi ta ch cn cp nht vic ging dy ta phi m bo khng gy nh hng ti cc gio vin khng tham gia ging dy v ngc li. Nh vy thng tin lu tr lc quan h ny khng chc chn. 1.2. Phn r T mt lc quan h km cht lng ban u cng vi tp ph thuc hm ca n ta tun theo mt nguyn tc no phn r thnh nhng lc quan h cht lng hn. v d: Phn r lc quan h GIANGDAY thnh hai lc TKB v GV TKB(MONHOC, SOTIET, LOP) GV(LOP, GV, HOCVI, DC) Tnh trng d liu ca hai lc trn nh sau: TKB=  EMBED Equation.3  GV= EMBED Equation.3  MONHOCSOTIETLOP CSDL90TH1CSDL90TH2CTDL75TH1CTDL75TH2LOPGVHOCVIDCTH1N.V.ACNQ1TH2N.V.ACNQ1TH1N.V.BTSQ2TH2N.V.BTSQ2 Sau y l hai rc ri xy ra tr li cu hi Cho bit thng tin ca gio vin dy CSDL ca TH1 ta phi kt ni t nhin hai quan h TKB v GV. Kt qu nh sau: MONHOCSOTIETLOPGVHOCVIDCCSDL90TH1N.V.ACNQ1CSDL90TH1N.V.BTSQ2CSDL90TH2N.V.ACNQ1CSDL90TH2N.V.BTSQ2CTDL75TH1N.V.ACNQ1CTDL75TH1N.V.BTSQ2CTDL75TH2N.V.ACNQ1CTDL75TH2N.V.BTSQ2Ta thy rng c ti hai gio vin dy mn CSDL ca lp TH1 trong khi thng tin ban u ch c N.V.A ( Vn ny gi l phn r khng bo ton thng tin Xt ph thuc hm trn lc phn r: TKB(MONHOC, SOTIET, LOP) MONHOC ( SOTIET GV(LOP, GV, HOCVI, DC) GV ( HOCVI, DC T hai ph thc hm trn ta khng th suy ra c ph thuc hm MONHOC, LOP (GV. Nh vy, hai ph thuc hm trn khng m bo kim tra cc rng buc ton vn do 3 ph thuc hm ban u gy ra nn. ( Vn ny gi l phn r khng bo ton ph thuc hm. Sau y, ta s xt cc quy tc phn r sao cho khng vi phm hai vn trn. Phn r bo ton thng tin Cho lc quan h Q. Ta c nh ngha sau: Tp {Q1,Q2,,Qn} l mt phn r ca Q nu:  EMBED Equation.3  Mt cch tng qut TQ l mt quan h ca Q th:  EMBED Equation.3  Phn r thng tin trn bo ton thng tin nu:  EMBED Equation.3  iu kin phn r bo ton thng tin Cho lc quan h Q v F l tp ph thuc hm nh ngha trn Q. Q1 v Q2 l mt phn r bo ton thng tin trn Q nu tho mt trong hai ph thuc hm sau:  EMBED Equation.3  Hoc  EMBED Equation.3  V vy nu  EMBED Equation.3  th phn r sau s bo ton thng tin Q1(XY) Q2(R-Y) Tht vy, v Q1 xc nh c X(Y v  EMBED Equation.3  do  EMBED Equation.3  V d: Lc GIANGDAY nu phn r thnh hai lc sau th bo ton thng tin Q1(MONHOC, SOTIET, LOP, GV) Q2(GV, HOCVI, DC) V  EMBED Equation.3  m GV (HOCVI,DC Liu mt php tch c kt ni khng mt mt thng tin hay khng i vi tp cc ph thuc hm c kim tra qua thut ton sau y. Thut ton 1.2.1. Kim tra php tch kt ni khng mt mt thng tin. Vo: Lc quan h R = {A1,. .., An}, tp cc ph thuc hm F v php tch ( = (R1,. .., Rk). Ra: Kt lun php tch p c kt ni mt mt thng tin hay khng?. Phng php : - Thit lp mt bng vi n ct v k hng ; ct th j ng vi thuc tnh Aj, hng th i ng vi lc Ri. - Ti hng i v ct j in k hiu aj nu Aj EMBED Equation.3Ri, nu khng in k hiu bij. - By gi xem xt n cc ph thuc hm trong F p dng cho bng va thit lp c. mi ln xt mt ph thuc hm (XEMBED Equation.3Y) EMBED Equation.3F, tm cc hng c gi tr bng nhau tt c cc ct trong X (chng hn c hai hng t1, t2 m t1[X] = t2[X]) th lm bng cc gi tr tt c cc ct ca hai hng trong Y (chng hn t1[Y]=t2[Y]). Ch khi lm bng gi tr trn mi ct trong Y, nu mt trong hai gi tr trn hai hng l aj th u tin lm bng chng bng k hiu aj. Nu khng th lm bng chng bng mt trong hai k hiu bij. Tip tc p dng cc ph thuc hm cho bng (k c vic lp li ph thuc hm c p dng) cho ti khi khng cn p dng c na. Xem xt bng kt qu: Nu xut hin mt hng gm cc k hiu a1,a2,. .., an (hng cha ton a) th php tch kt ni khng mt mt thng tin. Ngc li th php tch kt ni mt mt thng tin (lossy join). V d 4.9: Xem xt v d quan h cung cp trn S (SNAME, ADD, PRO, PRICE) c tch lm hai quan h S1(SNAME, ADD) v S2(SNAME, PRO, PRICE) Vi tp ph thuc hm : SNAMEEMBED Equation.3ADD, SNAME, PROEMBED Equation.3PRICE. Bng ban u c thit lp nh sau. SNAMEADDPROPRICEa1 a1a2 b22b13 a3b14 a4p dng ph thuc hm SNAMEEMBED Equation.3ADD cho hai hng ca bng. Hai hng lm bng a2. Bng kt qu l: SNAMEADDPROPRICEa1 a1a2 a2b13 a3b14 a4Bng kt qu c dng th hai c cc gi tr ton l a, do n l php tch kt ni khng mt mt thng tin. nh l 4.4 Thut ton 4.3 xc nh chnh xc mt php tch kt ni c hay khng mt mt thng tin. T nh l trn d dng kim tra trng hp tch thnh hai lc con qua nh l sau: nh l 4.5 Nu ( = (R1, R2) l mt php tch ca lc quan h R v F l tp ph thuc hm th ( l tch khng mt mt thng tin i vi F khi v ch khi R1EMBED Equation.3R2 EMBED Equation.3R1 - R2 hoc R1EMBED Equation.3R2EMBED Equation.3R2 - R1. Ch rng cc ph thuc hm nu trn khng nht thit phi thuc tp ph thuc hm F ban u, ch cn chng thuc F+. Chng minh Bng ban u c thit lp nh sau. R1EMBED Equation.3R2R1 - R2R2 - R1Hng cho R1 Hng cho R2aa... a aa... aaa... a bb... bbb... b aa.... aTa thy ngay Nu R1EMBED Equation.3R2 EMBED Equation.3R1 - R2 hoc R1EMBED Equation.3R2EMBED Equation.3R2 - R1 th R1EMBED Equation.3R2 EMBED Equation.3R1 - R2 nn dng hai ca bng kt qu ton a. Hoc R1EMBED Equation.3R2 EMBED Equation.3R2 - R1 nn dng mt ca bng kt qu ton a. Do php tch kt ni khng mt mt thng tin. Ngc li nu php tch trn c kt ni khng mt mt thng tin th bng kt qu c dng mt hoc dng hai ton a. Nu dng mt ton a th thy ngay R1EMBED Equation.3R2 EMBED Equation.3R2 - R1 Cn nu dng hai ton a th thy ngay R1EMBED Equation.3R2 EMBED Equation.3R1 - R2. V d 4.10 : R=ABC v F={ AEMBED Equation.3B }, ( = (R1,R2), R1= AB, R2= AC. Ta c A = R1EMBED Equation.3R2 B = R1 - R2 M AEMBED Equation.3B tc l R1EMBED Equation.3R2 EMBED Equation.3R1 - R2 nn php tch trn kt ni khng mt mt thng tin. Phn r bo ton ph thuc hm Theo phn trn mt php tch cn phi c c tnh kt ni khng mt mt thng tin v n cho php khi phc li mt quan h t cc hnh chiu ca n. Mt c tnh tnh quan trng khc ca php tch ( = (R1,R2,. .. Rk ) ca lc quan h R l c th suy ra c tp cc ph thuc hm F ca R t cc hnh chiu ca F trn cc Ri. 1) Khi nim php tch bo ton ph thuc hm ( Cho lc quan h R, F l tp cc ph thuc hm V ( = (R1, R2, ..., Rk) l mt php tch. Chiu ca tp ph thuc hm F trn mt tp Ri k hiu EMBED Equation.3Ri(F) gm nhng ph thuc hm X-->Y ca F+ m X-->Y nm gn trong Ri ( XY ( Ri ) <=>EMBED Equation.3Ri(F) ={X--> Y(F+ ( XY( Ri} Quy c Cho R l mt lc quan h F l tp cc ph thuc hm ca R K hiu W = Gi l mt s quan h (SQH) V d 4.11: W= -SQH R={A,B,C,D} F={A-->B, B-->C, C-->D, D-->A} ( = (R1, R2, R3) R1={A, B}, R2={B, C},R3 ={C, A} Suy ra: ( R1 (F) ={ A-->B, B-->A} ( R2 (F) ={ B-->C, C-->B} ( R3 (F) ={ C-->D, D-->C} Vy cho W= ( Ta ni php tch ( =(R1, R2, ..., Rk) Bo ton ph thuc hm nu F ~ EMBED Equation.3EMBED Equation.3Ri(F) V d trn c phn r bo ton ph thuc hm V F ~ EMBED Equation.3EMBED Equation.3Ri(F) ={A-->B, B-->A, B-->C, C-->B, D-->C, C-->D}=G Tht vy Hin nhin G ( F+ Tpcm F ( G+ Tht vy D --> A (F m D --> C ( G+ C -->B ( G+ B-->A( G+ =>D-->A ( G+ Chng minh tng t cho cc ph thuc hm khc. Vy php tch trn c bo ton ph thuc hm. Ch : Mt php tch c th c kt ni khng mt mt thng tin ng vi mt tp ph thuc hm F nhng khng bo ton F. Cng c php tch c th bo ton F nhng khng c tnh cht kt ni khng mt mt thng tin. V d 4.12: Cho LQH R={C, S, Z} F= {CS ( Z, Z( C} ( =(SZ, CZ) c kt ni khng mt mt thng tin v (SZ EMBED Equation.3CZ) ( (CZ - SZ ) nhng khng bo ton ph thuc hm. Ngc li Cho R= abcd F={A( B, C( D} v ( =(AB, CD) c ni mt mt thng tin nhng bo ton ph thuc hm. 2) Thut ton kim tra php tch bo ton ph thuc hm Vo: W= - SQH ( =(R1, R2, ...,Rk) l mt php tch Ra: Khng nh php tch ( c bo ton ph thuc khng? Phng php: tng ( bo ton ph thuc hm <=> F ~ EMBED Equation.3EMBED Equation.3(F) =G Hin nhin G ( F+ Vy ( c bo ton ph thuc hm <=> F ( G+ <=>( X-->A (F th X-->A (G+ <=>XG+ ( A Vy thut ton xt ( c bo ton ph thuc khng? b1: Tnh G = EMBED Equation.3EMBED Equation.3Ri(F) b2: ( X-->Y (F xt xem XG+ ( Y khng? Nu : ( X-->Y (F ta u c XG+ ( Y th kt lun: ( c bo ton ph thuc hm. Nu ( X-->A (F m XG+ khng cha Y th kt lun: ( khng bo ton ph thuc hm. V d 4.13: R= {A,B,C,D,E} F={AB->C, CD->E} P=(R1, R2, R3) R1 ={A,B}; R2 ={C,D}; R3 ={D,E}; B1: Tnh G={ AB->A, AB->B, CD->C, CD->D} B2 : AB->C ( G+ ? Ta c: {AB}G+=EMBED Equation.3{AB} khng cha C. Vy php tch trn khng bo ton ph phuc hm. 1.3.Cc dng chun 1.3.1. Chun ho lc quan h Nu mt lc quan h thit k khng tt s gy ra nhng d thng d liu nh d tha d liu v do vic cp nht d liu (qua php tnh chn, loi b v thay i d liu). trnh d thng d liu lc quan h cn thit phi c bin i thnh cc dng ph hp. Qu trnh c xem l qu trnh chun ho lc quan h. Lc quan h c chun ho l lc quan h trong min ca mi thuc tnh ch cha nhng gi tr nguyn t (atomic) tc l khng phn nh c na v do mi gi tr trong quan h cng l nguyn t. C th mi min khng ly gi tr kiu nh bn ghi hay a tr. Ngc li gi l lc quan h khng chun ho. Qu trnh chun ho mt lc quan h c th thnh mt hoc nhiu lc quan h chun ho khc v c kt ni khng mt mt thng tin. Lc quan h R l chun ho th quan h r ca n cng l chun ho. V d 1.3.1a: GVMHPHONGGVMHPHONGA Csdl414, H3A ACSDL CSDL414 H3BCTDL408BCTDL408Quan h khng chun ho Quan h chun ho V d 1.3.1b: MaGVTenGVDiachiMaGVTenGVHuyenTinhHuyenTinh1AXY1AXY2BWZ2BWZQuan h khng chun ho Quan h chun ho Trc khi m t chi tit cc dng chun cn thit a ra mt s khi nim sau. Cho mt lc quan h R vi tp thuc tnh U={A1,..., An}. Thuc tnh AEMBED Equation.3U c gi l thuc tnh kho nu A thuc mt kho no ca R, ngc li A c gi l thuc tnh khng kho. V d 1.3.1.1: Cho lc quan h R trn tp thuc tnh U={A,B,C,D} vi cc ph thuc hm EMBED Equation.3, EMBED Equation.3 v EMBED Equation.3. R rng rng AB v BC l kho ca lc R. Khi A, B v C l thuc tnh kho, cn D l thuc tnh khng kho. nh ngha 1.3.1 Cho lc quan h R trn tp thuc tnh U={A1,.., An }. X v Y l hai tp thuc tnh khc nhau EMBED Equation.3 V EMBED Equation.3. Y ph thuc hm y (fully functional dependence) vo X nu Y ph thuc hm vo X nhng khng ph thuc hm vo bt k mt tp hp con thc s no ca X. 1.3.2. Cc dng chun Trong l thuyt ban u Codd a ra c ba dng chun ca quan h : Dng khng chun ho EMBED Equation.3Dng chun th nht (First Normal Form, vit tt l 1NF) EMBED Equation.3Dng chun th hai (Second Normal Form, vit tt l 2NF) EMBED Equation.3Dng chun th ba (Third Normal Form, vit tt l 3NF) Sau ny cn c nhiu dng chun khc nh: BCNF, 4NF, 5NF,...Dng chun th 1 (first normal form) nh ngha 1: Mt lc quan h R vi tp thuc tnh U c gi l dng chun mt (1NF) nu min ca mi thuc tnh trong U ch cha nhng gi tr nguyn t (atomic). nh ngha ny cho thy: Bt k mt lc quan h chun ho no cng dng 1NF v quan h r ca n cng dng chun 1. Quy c : Mt lc quan h nu khng nu r n khng l 1NF th coi nh n 1NF. Dng chun th 2 -Second normal form ( dng ny ch tn ti vi ngha lch s) nh ngha 2: Mt lc quan h R vi tp thuc tnh U c gi l dng chun th hai nu n dng chun th nht v mi thuc tnh khng kho ca R u ph thuc hm y vo kho. Hay ni mt cch khc mi thuc tnh khng kho ca n khng ph thuc hm vo mt tp con thc s no ca kho. V d 1.3.2a: * Cho lc quan h R(SAIP) vi ph cc thuc hm EMBED Equation.3 v EMBED Equation.3. R khng l 2NF. Tht vy: t EMBED Equation.3, Y=S. Ta c A l thuc tnh khng kho v R ch c 1 kho l SI. M Y ( X v EMBED Equation.3. Nh vy A khng ph thuc hm y vo kho. * Cho lc quan h R(SAIP) -1NF vi ph thuc hm EMBED Equation.3, EMBED Equation.3. R l 2NF. Tht vy: Ta c A, P l cc thuc tnh khng kho. R ch c 1 kho l SI v khng c tp con no ca SI xc nh hm A v P tc l S hoc I khng khng xc nh hm Av P. Nh vy A, P ph thuc hm y vo kho SI. Dng chun th ba (third normal form) nh ngha 3: Mt lc quan h R vi tp ph thuc hm F c gi l dng chun th 3 (3NF) nu mi ph thuc hm X -> A ng trong R v A (X th hoc X+ = U hoc A l thuc tnh kho. Hay ni mt cch khc khng tn ti mt ph thuc hm dng XEMBED Equation.3A vi X ( U ; A (X v A khng l thc tnh kho m X+ (U. V d 1.3.2b: *Cho lc quan h R(CSZ) -1NF v ph thuc hm EMBED Equation.3. Trong lc ny ch c mt kho l CS nn n l kho chnh, thuc tnh khng kho l Z. Thy ngay R l 3NF. * Cho lc R(SIDM) -1NF v cc ph thuc hm SIEMBED Equation.3D v SDEMBED Equation.3M. y ch c mt kho chnh l SI. Thuc tnh khng kho l D, M. R rng R 2NF nhng khng phi 3NF (v SDEMBED Equation.3M). Ch : Nu R khng cha thuc tnh khng kho th R dng 3NF. Dng chun Boye - Codd (Boye - Codd Normal form ) nh ngha 4: Mt lc quan h R vi tp ph thuc hm F c gi l dng chun Boye - codd (BCNF) nu XEMBED Equation.3A tho trn R v AEMBED Equation.3 X th X+ = U. Hay ni mt cch khc khng tn ti mt ph thuc hm dng XEMBED Equation.3A vi X ( U v A (X m X+ (U. T nh ngha ta thy ngay nu R BCNF th R 3NF. V trong 3NF ch cm cc thuc tnh A khng kho ph thuc vo tp X ( U m A (X v X+ (U, cn trong BCNF cm tt c cc thuc tnh A (X ph thuc vo tp X ( U m X+ (U. V d 1.3.2c: Cho R(CSZ) v tp th thuc hm EMBED Equation.3, Z( C r rng R khng BCNF m 3NF v ZEMBED Equation.3C nhng Z khng phi l kho ca R v C l thuc tnh kho. T v d ny thy rng mt lc quan h c th l 3NF nhng cha chc phi l BCNF. Nhng mt lc quan h l BCNF th n l 3NF. 2.Chun ho lc c s d liu 2.1. Phng php phn r Thut ton phn r lc quan h thnh BCNF kt ni khng mt mt thng tin Trc khi a ra thut ton phn r lc quan h thnh BCNF, cn kho st mt s tnh cht sau: B 4.6 a) Cho R l mt lc quan h vi tp ph thuc hm F. Gi EMBED Equation.3(R1, R2, . .., Ri ,..., Rk) l mt php tch kt ni khng mt mt thng tin ca R i vi F v EMBED Equation.3= (S1, S2) l php tch c kt ni khng mt mt thng tin ca Ri ng vi Fi l hnh chiu ca F trn Ri. Khi php tch ca R thnh (R1, . .., Ri-1, S1, S2, Ri+1,..., Rk) cng c kt ni khng mt mt thng tin i vi F. b) Mi lc c hai thuc tnh hoc tp ph thuc hm F = ( u c dng BCNF. c) Nu R khng c dng BCNF th c th tm c cc thuc tnh A v B trong U sao cho (U- AB) EMBED Equation.3A. Ph thuc hm (U- AB) EMBED Equation.3B c th cng ng trong trng hp ny. Chng minh: i vi trng hp (b), gi AB l lc ny. lc ny ch 2 ph thuc hm khng tm thng c th ng l AEMBED Equation.3B hoc B EMBED Equation.3A, nu khng c ph thuc hm no ng th chc chn AB l BCNF. Nu ch A EMBED Equation.3B ng th A l mt kho, v th khng c vi phm BCNF. Nu ch B EMBED Equation.3A ng th B l mt kho v nu c hai ng th c A v B u l kho v th khng c vi phm BCNF. i vi trng hp (c), gi s c mt vi phm X EMBED Equation.3A trong R th th phi c thuc tnh B khng thuc XA, nu khng X l mt kho bao hm v X EMBED Equation.3A khng phi l vi phm, v X ( ( U-AB) v X EMBED Equation.3A nn (U-AB) EMBED Equation.3A. Thut ton 4.4: Tch mt lc quan h thnh BCNF c ni khng mt mt thng tin Vo: Lc quan h R v tp ph thuc hm F. Ra: Php tch ca R c kt ni khng mt mt thng tin sao cho mi lc quan h trong php tch u BCNF i vi php chiu ca F trn lc . Phng php: Cu trc php tch ( trn R theo phng php lp lin tip. Ti mi bc php tch ( l bo m khng mt mt thng tin i vi F. Bc u: ( ch bao gm R v tp ph thuc hm l F Cc bc tip: Nu S l mt lc thuc (, S cha BCNF, chn XEMBED Equation.3A l ph thuc hm trong khng tho trn BCNF, tc l X khng cha kha ca S v A EMBED Equation.3X. Thay th S trong ( bi S1 v S2 vi S1 = XA v F1 =(S 1(F), S2 = S - A v F2 = (S 2(F) Theo nh l 4.5, php tch S thnh S1 v S2 c kt ni khng mt mt thng tin i vi tp ph thuc hm trn S v S1 EMBED Equation.3S2 = X, XEMBED Equation.3S1 - S2 = A. Theo b 4.6 phn (a), ( c thay S bng S1 v S2 c ni khng mt mt thng tin. Mi lc S1, S2 u c s thuc tnh t hn S. Qu trnh tip tc vi (S1, F1) v (S2, F2) cho ti khi tt c cc lc u BCNF. Ch rng ti mi thi im ( lun m bo khng mt mt thng tin, v rng ( ban u l R, m bc thay i ( u bo ton tnh cht . B 4.7 Cho tp ph thuc hm F trn R. Nu chiu F trn R1 ( R c F1, ri li chiu F1 trn R2 ( R1 c F2 th F2 =(R2(F) Theo b trn, nu ta phn r cc lc quan h theo thut ton 4.4 th thc s ta khng phi tnh nhng ph thuc hnh chiu trong qu trnh phn r, ngha l ta ch s dng tp ph thuc hm F cho, ly bao ng ca tp thuc tnh khi cn ch khng phi tnh tt c cc hnh chiu ca cc ph thuc hm. V d 4.19: cho lc R(CTHRSG), trong C: Gio trnh, T: Thy gio, H: Gi, R: Phng hc, S: Sinhvien, G: Lp. Tp cc ph thuc hm F nh sau: CEMBED Equation.3T : Mi gio trnh c mt thy dy HREMBED Equation.3C : Ch mt mn hc (gio trnh) mt phng hc ti mt thi im. HTEMBED Equation.3R : Ti mi thi im mi thy gio ch c th dy mt phng hc. CSEMBED Equation.3G : Mi sinh vin ch mt lp hc trong mi gio trnh HSEMBED Equation.3R : Mi sinh vin ch c th mt phng hc ti mi thi im. Kho ca R l HS. Tch lc R thnh cc lc BCNF. Xt CSEMBED Equation.3G cho R. vi phm iu kin BCNF v CS khng cha kho. Do vy, dng thut ton 4.3 tch R thnh R1(CSG) v R2(CTHRS). Bc tip cn tnh F+ v chiu xung R1 v R2, sau kim tra cc lc BCNF cha. C th biu din qu trnh tch qua s sau:  Php tch cui cng c R = (R1, R21,, R221, R222). 2.2.Phng php tng hp Php tch mt lc quan h thnh 3NF c bo ton ph thuc hm. Khng phi lc no cng tch c mt lc quan h thnh dng BCNF m vn bo ton c cc ph thuc hm. Tuy nhin lun c th tm c php tch bo ton ph thuc hm thnh dng chun th 3 nh thut ton sau: Thut ton 4.5: Tch mt lc quan h thnh 3NF Vo: Lc quan h R, tp cc ph thuc hm F (khng lm mt tnh tng qut gi s rng n l ph ti thiu). Ra: Php tch ca R bo ton cc ph thuc hm sao cho mi lc con u 3NF ng vi hnh chiu ca F trn lc . Phng php: a) Loi b tt c cc thuc tnh ca R nu cc thuc tnh khng lin quan n mt ph thuc hm no ca F (hoc v tri, hoc v phi). V nguyn tc mi thuc tnh nh th u c th to ra mt lc . b) Nu c mt ph thuc hm no ca F m lin quan ti tt c cc thuc tnh ca R th kt qa ra chnh l R. c) Ngoi ra, php tch ( a ra cc lc gm cc thuc tnh XA cho ph thuc hm XEMBED Equation.3A thuc F, nu XEMBED Equation.3A1, XEMBED Equation.3A2,..., XEMBED Equation.3An thuc F th thay th lc XA1A2...An cho cc lc XAi(1<=i<=n). V d 4.20: Cho lc quan h R(CTHRSG) vi tp ph thuc hm ti thiu CEMBED Equation.3T,HREMBED Equation.3C,HTEMBED Equation.3R, CSEMBED Equation.3G v HSEMBED Equation.3R dng cho thut ton 4.5 cho tp lc dng chun th 3: R1(CT), R2(CHR), R3(HRT), R4(CGS) v R5(HRS). nh l 4.6 Thut ton 4.5 to ra mt php tch bo ton cc ph thuc hm v cc lc ch dng chun 3. Php tch mt lc quan h thnh 3NF c bo ton ph thuc hm v kt ni khng mt mt thng tin. Nh bit, c th tch mt lc quan h R bt k thnh mt tp cc lc ( = (R1, R2,. .., Rn) sao cho ( c kt mi khng mt mt thng tin v mi Ri c dng BCNF (v th n dng 3NF). Ta cng c th tch R thnh ( = (S1, S2,. .., Sm) sao cho ( bo ton tp cc ph thuc hm F v mi S i c dng 3NF. Liu c th tm c mt php tch thnh 3NF m vn c c hai c tnh bo ton ph thuc hm v kt ni khng mt mt thng tin? Ta ch cn ni vo ( mt lc quan h X trong X l kho ca R nh nh l sau s trnh by. nh l 4.7 Gi ( l mt php tch ca R thnh 3NF c xy dng bng thut ton 4.5 v X l kho ca R. Th th ( = ( ({X} l mt php tch ca R m tt c cc lc quan h u c dng 3NF v php tch ny c c tnh bo ton ph thuc hm v kt ni khng mt mt thng tin. Chng minh: D dng chng minh c rng vi bt k vi phm 3NF no trong X cng khng nh c rng mt tp con thc s ca X xc nh hm X v nh th xc nh c R, do vy X khng th l kho trong trng hp ny. Nh vy X v cc lc trong ( phi c dng 3NF. R rng ( bo ton ph thuc v ( bo ton ph thuc chng minh ( c kt ni khng mt mt thng tin, ta xy dng cc bng v p dng thut ton 4.2, ta c th chng minh rng ton b hng ca X s tr thnh a nh sau. Xt th t A1, A2,. .., Ak l th t khi thm cc thuc tnh ca R-X vo X+ trong thut ton 4.1 chc chn rng cui cng th tt c cc thuc tnh s c thm vo bi v X l kho. Ta s chng minh bng quy np trn i rng ct tng ng vi Ai trong hng ca X c t bng a trong php kim tra ca thut ton 4.2. Bc c s i = 0 hin nhin ng. Gi s kt qu ng vi i-1. Th th Ai c thm vo X+ do c mt ph thuc hm Y( Ai trong : Y(X({A1, A2,..., Ai-1}. Th th YAi thuc ( v X ging nhau Y (tt c u l a) sau khi cc ct cho A1, A2,..., Ai-1 hng X c t l a. V vy cc hng ny c bin i thnh ging nhau thuc tnh Ai trong khi thc hin thut ton 4.2. Bi v hng YAi c k hiu ai nn hng X cng th. V d 4.21: Ta c th ly hp ca lc CSDL ca CTHRSG c to ra trong v d 4.20 vi kho SH, thu c php tch c kt ni khng mt mt thng tin v bo ton ph thuc hm. May mn SH li l tp con ca HRS l mt trong nhng lc quan h c trong php tch v th c th loi b SH v chp nhn lc CSDL ca v d 4.20, ngha l (CT, CHR, THR, CSG, HRS). Cu hi v bi tp chng 5 Bi 5.1: Nu cc nh ngha dng chun 1NF, 2NF, 3NF, BCNF cho v d. Bi 5.2: Trnh by gii thut kim tra php tch SQH c kt ni khng mt tin ? v trnh by thut ton kim tra php tch SQH bo ton ph thuc hm, cho v d. Bi 5.3: Trnh by thut ton tch SQH bo ton ph thuc hm v dng 3NF, cho v d. Bi 5.4: Trnh by thut ton tch SQH c kt ni khng mt tin v v dng BCNF, cho v d. Bi 5.5: Trnh by thut ton tch SQH c kt ni khng mt tin v bo ton ph thuc hm v dng 3NF, cho v d. Bi 5.6: Cho lc quan h R v tp cc ph thuc hm a. F= {AB -> E, AG -> I, BE -> I, E -> G, GI -> H} trn R Chng minh rng: AB -> GHI. b. F= {AB -> C, B -> D, CD -> E, CE -> GH, G -> A} trn R Chng minh rng: AB -> E v AB -> G. Bi 5.7: Cho s quan h W = vi R = {A, B, C,D}v F= {A -> B, A -> C}. Ph thuc hm no trong dy sau c suy dn t F: A -> D C -> D BC -> A A -> BC F= {A -> B, BC -> D}. Ph thuc hm no trong dy sau c suy dn t F: C -> D A -> D AD -> C BC -> A B -> CD Bi 5.8: Cho s quan h W = . Cc php tch sau c kt ni mt tin khng ?. Vi a. R = {A, B, C}v F = {A -> B} P = (R1, R2), R1= {A, B}, R2= {A, C}. b. R = {A, B, C}v F = {A -> B, A -> C } P = (R1, R2), R1= {A, B}, R2= {A, C}. Bi 5.9: Cho s quan h W = . Cc php tch sau c kt ni mt tin khng ?. Vi a. R = {A, B, C, D, E, H, I, K, L}v F = {AB -> D, DE -> H, IK -> L, LB -> C} P = (R1, R2, R3), R1 = {A, B, C}, R2 = {C, D, E, H}, R3 = {E, H, I, K, L}. b. R = ( ABCDEF) F= {AB -> C, C -> D, D -> E, DE -> F } P = (R1, R2, R3, R4), R1= {A, B, C}, R2= {C, D}, R3 = {D, E}, R4 = {D, E, F}. Bi 5.10 a)Cho s quan h sau: W= < R, F> Trong : R = ( ABCD) F= {AB -> C, D -> B, C -> ABD } V php tch ( =(ABC, BCD) Hy kim tra xem php tch trn c kt ni mt mt thng tin khng?, c bo ton ph thuc hm khng?. b. W= < R, F> Trong : R = ( ABCDEG) F= {AB -> C, C -> A, BC -> D, ACD -> B, D -> EG, BE -> C, CG -> BD, E -> G } V php tch ( =(ABC, BCD, DEG) Hy kim tra xem php tch trn c kt ni mt mt thng tin khng?. C bo ton ph thuc hm khng?. Bi 5.11: Hy tch s quan h W= < R, F> Trong : R = ( ABCDE ) F= {AB -> C, C -> D, D -> E } thnh cc lc quan h dng BCNF c kt ni khng mt mt thng tin. 3NF bo ton ph thuc hm v 3NF c kt ni khng mt mt thng tin v bo ton ph thuc hm. Bi 5.12: Cho s quan h W = .Trong : R = (masv, mamon, diem, hoten, ngaysinh, gioi, diachi, quequan) F = {masv, mamon --> diem, masv --> hoten, ngaysinh, gioi, diachi , quequan} R1 = {masv, mamon, diem} R2 = {masv, hoten, ngaysinh, gioi, diachi, quequan} ( = (R1,R2) Php tch trn c kt ni mt mt thng tin khng?, c bo ton ph thuc hm khng?. Bi 5.13: Cho S quan h QLCH = . Trong : R = ( Makh, Tenkh, Diachi, Gioitinh, Dienthoai, Mamh,Tenmh, mau, Dongia, Dvtinh, Soluong, MuaBan, NgayMB ). X = { Tenkh, Diachi, Gioitinh, Dienthoai } Y = { Tenmh, mau, Dongia, Dvtinh } Z = { Soluong, MuaBan, NgayMB } F= { Makh -> X, Mamh -> Y } Tm kho ca lc quan h R bng nh ngha kho v bng gii thut tm kho. Xt xem R c dng chun 2NF, 3NF, BCNF khng ?. Tch R thnh cc lc quan h dng BCNF c kt ni khng mt mt thng tin. 3NF bo ton ph thuc hm v 3NF c kt ni khng mt mt thng tin v bo ton ph thuc hm.. Tm kho ca mi lc quan h va nhn c t php tch trn. Bi 5.14: Cho S quan h QLDiem = . R = ( MaSV, TenSV, Diachi, Gioitinh, Lop, MaMon,TenMon, soHT, DiemthiL1, DiemthiL2, DiemthiL3). X = { TenSV, Diachi, Gioitinh, Lop } Y = { TenMon, soHT } Z = { DiemthiL1, DiemthiL2, DiemthiL3} MA = { MaSV, MaMon} F= { MaSV -> X, MaMon -> Y, MA -> Z } Tm kho ca lc quan h R bng nh ngha kho v bng gii thut tm kho. Tm ph ti thiu F ca F Xt xem R c dng chun 2NF, 3NF, BCNF khng ?. Tch R thnh cc lc quan h dng BCNF c kt ni khng mt mt thng tin. 3NF bo ton ph thuc hm v 3NF c kt ni khng mt mt thng tin v bo ton ph thuc hm. Tm kho ca mi lc quan h va nhn c t cc php tch trn. Chng VI Ti u ho cu hi Ti u ha cu hi l qu trnh la chn phng php sao cho khi thc hin cc cu hi truy vn c hiu qu nht, c th nh gi c cc kh nng x l cu hi t nhiu chin lc khc nhau, c bit l cho nhng cu hi phc tp. Mt phng php khi thc hin c chi ph thp, tc l ti u v thi gian truy xut thng tin v ti u v khng gian lu tr m vn bo m c tnh c lp v ton vn d liu. 1. Cc nguyn tc tng qut ti u ho cu hi: 1.1. Cc nguyn tc tng qut - u tin thc hin cc php chiu v chn, nhm gii hn khi lng d liu trung gian. Gim chi ph truy nhp b nh. - Trc khi phi thc hin php tch cc, hy tm chin lc truy nhp tt nht vo CSDL. V d nh s dng cc php sp xp, hoc chn ch s trn thnh phn tham gia vo tch cc. - Thc hin cc php kt ni cn bng chi ph s r hn nhiu so vi chi ph thc hin php tch cc. - Nhm cc php ton chn v chiu lin tip thnh mt php ton duy nht. - Nhm cc php tch v chiu lin tip thnh mt php ton duy nht. Trong khi thi thc hin php tch c th gii hn chi ph thc hin bng php chiu. - Tm biu thc chung trong mt biu thc. Nu kt qu l mt quan h khng ln lm nhng tn xut xut hin nhiu ln, nn c biu thc con chung. - nh gi s b trc khi thc hin cu hi. S php ton thc hin, tng chi ph thc hin: thi gian, b nh ... 1.2.Biu thc tng v cc quy tc Biu thc trong ngn ng i s quan h c cc hng thc l bin quan h (relation variables) R1,R2,,Rn; cc quan h hng c xc nh nh l mt nh x t cc k b ca cc quan h (r1, r2, ,rk) trong ri l quan h trn lc ri v thay th ri vo Ri khi nh gi biu thc. Hai biu thc E1 v E2 c gi l tng ng, vit tt l  EMBED Equation.3 , nu chng biu din cng mt nh x, ngha l, nu chng ta thay th cng cc quan h cho tn cc lc tng ng hai biu thc E1 v E2, th chunngs s cho ra cng mt kt qu. 1.2.1. Cc quy tc lin quan ti php kt ni v tch cc ( Quy tc giao hon ca php kt ni v tch cc Nu E1 v E2 l hai biu thc quan h v F l iu kin trn cc thuc tnh ca E1 v E2 th:  EMBED Equation.3  // Tnh giao hon ca php kt ni  EMBED Equation.3  // Tnh giao hon ca php kt ni bng  EMBED Equation.3  // Tnh giao hon ca tch cc Ch : Nu quan nim quan h l tp cc b (c th t thuc tnh c nh) th php kt ni, kt ni t nhin v tch cc khng th giao hon c v cc thuc tnh trong quan h kt qu b thay i. ( Quy tc kt hp ca php kt ni v tch cc Nu E1, E2, E3 l cc biu thc quan h: F1, F2 l iu kin th:  EMBED Equation.3   EMBED Equation.3  1.2.2. Cc quy tc lin quan ti php chn v php chiu Dy cc php chiu c th t hp li thnh mt php chiu, biu din theo cc trng hp sau: ( Dy cc php chiu (E[B1B2& Bm])[A1A2& An] a" E[A1A2...An] y, cc thuc tnh A1, ...,An phi nm trong tp cc thuc tnh B1,...,Bm. Ng ngha ca vic bin i tng ng ny l: Nu thc hin mt php chiu biu thc quan h E trn tp cc thuc tnh B, ri sau thc hin tip php chiu trn tp con cc thuc tnh  EMBED Equation.3  ca quan h va tm c, th kt qu ca dy php chiu ny hon ton tng ng vi mt php chiu biu thc quan h E trn tp thuc tnh A. Tng t, dy cc php chn c th t hp thnh mt php chn kim tra tt c cc iu kin cng mt lc v c biu din nh sau: ( Dy cc php chn: (((E: (f1)):f2):...):fn a" E: (f1^f2 ... ^fn) Vic ln lt thc hin cc php chn trn quan h kt qu ca mt php chn trc i vi biu thc quan h E l tng ng vi vic chn trn E cc b gi tr tho mn ng thi tt c cc iu kin chn f1, f2, ...fn. ( Giao hon php chn v php chiu (E[A1& An]: (f)) a" (E: (f))[A1...An] Mt cch tng qut hn nu iu kin chn f lin quan ti cc thuc tnh B1, ..., Bm m khng nm trong tp thuc tnh A1,...,An th: (E[A1...An]): (f) a" ((E[A1...AnB1...Bm]): (f))[A1...An] ( Giao hon php chn v tch cc: Nu tt c cc thuc tnh ca F l thuc tnh ca E1 th: (E1x E2): (f) a" (E1: (f))x E2 T d dng suy ra rng, nu F c dng f= f1 ^ f2 trong f1 ch lin quan ti cc thuc tnh ca E1; f2 ch lin quan ti cc thuc tnh ca E2, th c th s dng cc trng hp (,(,( c: (E1xE2): (f) a" (E1: (f1))x (E2: (f2)) Hn na nu f1 ch lin quan ti cc thuc tnh ca E1, nhng f2 lin quan ti cc thuc tnh ca c E1 v E2 th: (E1x E2): (f) a" ((E1: (f1)) x E2): (f2) ( Giao hon php chn v mt php hp Nu c biu thc  EMBED Equation.3  c th gi thit thm rng, cc thuc tnh ca E1 v E2 c cng tn nh ca E hoc t nht mi thuc tnh ca E l ph hp vi mt thuc tnh duy nht ca E1 v mt thuc tnh duy nht ca E2, Khi :  EMBED Equation.3  Nu tn cc thuc tnh ca E1 v hoc E2 khc vi tn thuc tnh ca E th trong f v phi ca cng thc trn cn thay i s dng tn cho ph hp. ( Giao hon php chn v mt php hiu tp hp (E1- E2): (f) a" (E1: (f))  (E2: (f)) Nh nu trong trng hp (, nu tn cc thuc tnh ca E1 v E2 l khc nhau th cn thay th cc thuc tnh trong f v phi biu thc tng ng tng ng vi E1. Ch rng, php chn (E2: (f)) c th l khng cn thit. Trong nhiu trng hp, vic thc hin php chn (E2: (f)) trc s c hiu qu hn l tnh ton trc tip vi E2 v kch c quan h lc s b i rt nhiu. Cc quy tc nu trn ni chung l y php chn xung trc php kt ni v php kt ni thng thc hin lu nh php tch Cc. Quy tc y php chn xung trc php kt ni suy ra t quy tc (, ( v(. Quy tc y php chiu xung trc php tch cc hoc php hp cng tng t nh quy tc ( v (. Ch l khng c phng php tng qut cho vic y php chiu xung trc php hiu cc tp hp. ( Giao hon mt php chiu vi mt php tch cc Gi E1, E2 l hai biu thc quan h, A1 An l tp cc thuc tnh trong B1, Bm l cc thuc tnh ca E1, cc thuc tnh cn li C1,,Ck thuc E2. Khi : (E1x E2) [A1An] a" E1[B1& Bm] x E2[C1& Ck] (Giao hon mt php chiu vi mt php hp  EMBED Equation.3  Nh nu trong lut (, nu tn cc thuc tnh ca E1 v hoc E2 l khc vi cc thuc tnh trong  EMBED Equation.3  th cn thay A1& An bn v phi bng cc tn ph hp 2.V d v mt thut ton ti u ho biu thc quan h Ti y c th p dng cc quy tc nu trong mc 1.2 c th ti u ho cc biu thc quan h. Biu thc ti u cc nguyn tc phi tun theo cc nguyn tc nu phn trn mc d rng cc nguyn tc khng c ngha l bo m ti u cho mi trng hp tng ng. Lu rng, lun lun y php chn v php chiu xung mc cng su cng tt trong cy biu din biu thc quan h nhm to nn mt dy cc php chn cng nh php chiu t c th t chc thnh mt php chn theo sau mt php chiu. Nhm cc php chn v php chiu li trong mt nhm thc hin trc cc php tnh hai ngi nh php hp, tch cc, hiu tp hp C mt s trng hp c bit xy ra khi mt php tnh hai ngi c cc hng thc cha php chn v hoc php chiu c p dng i vi l ca cy biu din biu thc. Khi cn xem xt cn thn tc ng ca php tnh hai ngi v mt s trng hp cn phi lin kt php chn hoc php chiu vi php hai ngi . Kt qu u ra (Output) ca thut ton l mt chng trnh bao gm cc bc sau: p dng ca php chn hoc mt php chiu n gin. p dng ca mt php chn v mt php chiu hoc p dng ca mt tch cc, php hp hoc php hiu tp hp cho hai hng thc m trc cc php chn hoc cc php chiu c p dng cho mt hoc c hai hng thc. Hy xt mt CSDL qun l th vin bao gm cc quan h sau y: SACH(Tn-sch, Tc-gi, NhXB, M-sch): l quan h v cc loi sch trong th vin. Nha-Xuat-Ban(Nh XB, a-ch, Thnh-ph): quan h v nh xut bn. Doc-Gia(Tn-G, ch-G, Tph-G, S-th):quan h v c gi. MUON-SACH(S-th, M-sch, Ngy-mn):quan h s theo di mn. lu tr thng tin v sch c th gi thit thm rng c mt khung nhn (VIEW) theo di cc sch c mn, TD-Mn, bao gm mt s thng tin b sung v sch c mn, l kt qu ca php kt mi t nhin ca quan h SACH, Doc-Gia v MUON-SACH, chng hn c xc nh qua biu thc quan h֠: ((SACH x Doc-Gia x MUON-SACH): (f)) [{S}] y: f::= (Doc-Gia.S-th= MUON-SACH.S-th) ^ (SACH.M-sch=MUON-SACH.M-sch). V S l tp thuc tnh: S= {Tn-sch, Tc-gi,NhXB, SACH.M-sch, Tn-G, ch-G, Tph-G, Doc-Gia.S-th, Ngy-mn}. Cu hi: Cho bit danh sch ca nhng cun sch cho mn trc ngy 27/03/1999 Biu thc quan h c vit nh sau: (TD-Mn: (Ngy-mn< 27/03/1999))[Tn-sch] Hnh cy ca biu thc trn c biu din bng hnh 6.1. Lu l, cc php ton nm pha di l cc php ton c thc hin trc cc php ton pha trn ca cy. [Tn-sch] //chiu ly ct tn sch   : (Ngy-mn<27/03/1999) //Chn sch cho mn trc ngy 27/03/199 [Tn-sch, Tc-gi, SACH.M-sch, Tn-G, ch-G,Tph-G, Doc-Gia.S-th, Ngy-mn] : ((SACH.M-sch=MUON-SACH.M-sch) ^ (MUON-SACH.S-th=Doc-Gia.S-th))  SHAPE \* MERGEFORMAT  X X SACH MUON-SACH Doc-Gia Hnh 6.1: Biu din cy ca biu thc hi Thay th cc gi tr f v S vo biu thc hi c c cy biu din ca biu thc quan h nh trong hnh 6.1 Bc th nht ca ti u l tch php chn f thnh hai php chn vi iu kin: Sach.M-sch= MUON-SACH.M-sch MUON-SACH.S-th=Doc-Gia.S-th By gi chng ta c 3 php chn. Cn y chng xung mc thp hn chng no cn c th c. php chn vi iu kin Ngy-mn <27/03/1999 c y xung di php chiu v hai php chn kia bng cch p dng cc quy tc (hoc lut) ( v(. php chn u c p dng cho tch cc (MUON-SACH x Doc-Gia) x SACH). V thuc tnh Ngy-mn trong php chn ch c quan h MUON-SACH nn c th thay thՠ: ((MUON-SACH x Doc-Gia) x SACH): (Ngy-mn <27/03/1999) bng biu thc: ((MUON-SACH x Doc-Gia): (Ngy-mn <27/03/1999)) x SACH V tip tc y xung na, cui cng ta c biu thc: (((MUON-SACH: (Ngy-mn< 27/03/1999)) x Doc-gia) x SACH) Nh vy y c php chn theo ngy mn sch ny xung su nh c th. By gi tip tc y php chn vi iu kin SACH.M-sch = MUON-SACH.M-sch xung mc thp nht nu c th. Khng th y php chn ny xung di tch cc v n lin quan ti mt thuc tnh ca quan h SACH v mt thuc tnh thuc quan h MUON-SACH. Do vy php chn: : MUON-SACH.S-th = Doc-Gia.S-th C th y xung p dng cho tch cc: (MUON-SACH x Doc-Gia): (: (Ngy-mn <27/03/1999)) Ch rng MUON-SACH.S-th l tn mt thuc tnh ca php chn: MUON-SACH: (Ngy-mn < 27/03/1999) Bc tip theo: T hp hai php chiu thnh mt php chiu l [Tn-sch] nh lut (. Kt qu c cho nh trong hnh 6.2. Sau p dng quy tc m rng ( thay thՠ: : (MUON-SACH.S-th= Doc-Gia.S-thS-th) v chiu [Tn-sch] nh dy php ton: [Tn-sch.SACH.M-sch.MUON-SACH.M-sch](1) : (SACH.M-sch= MUON-SACH.M-sch) (2) Ri chiu ly tn sch: [Tn-sach] (3) p dng quy tc ( thay th biu thc u tin, biu thc s (1), ca php chiu nh php chiu: [Tn-schTn-sch.SACH.M-sch] p dng cho quan h SACH v [MUON-SACH.M-sch] p dng cho hng thc pha tri ca tch cc trong hnh 6.2. php chiu cui v php chn c th p dng quy tc m rng ( c dy: [MUON-SACH.M-schM-sch, MUON-SACH.S-thS-th, Doc-Gia.S-th] (5) : (MUON-SACH.S-th= Doc-Gia.S-th) (6) [MUON-SACH.M-sch]M-sch] (7) Php chiu u, s (5) c phn tch chuyn xung tch cc nh quy tc ( Mt phn php chiu Doc-Gia.S-th xung hng thc Doc-Gia v l thuc tnh ca quan h ny. [Tn-Sch]    : (SACH.M-sch=MUON-SACH.M-sch) X (MUON-SACH.S-th= Doc-Gia.S-th) SACH X : (Ngy-mn <27/03/1999) Doc-Gia MUON-SACH Hnh 6.2 Cy vi t hp php chn v php chiu Phn cn li l php chiu ly 3 thuc tnh: [MUON-SACH.M-sch, MUON-SACH.S-thS-th, Ngy-mn] c y xung hng thc th MUON-SACH. Cc thuc tnh khng ph hp s b loi b. Biu din cy cui cng ca biu thc nh trong hnh 6.3 Nhm cc php tnh bng ng mi tn gin on. Mi tch cc c t hp vi php chn to thnh mt php kt ni bng nhau rt c hiu qu.c bit php chn trn quan h MUON-SACH v php chiu ca Doc-Gia ly thuc tnh s-th pha di l t hp vi tch cc. Th t thc hin ca cy biu thc trong cc hnh 6.1, 6.2 v 6.3 l t di ln: Nhm cc php ton nm pha di c thc hin trc cc php ton pha trn. [Tn-Sch]    : (SACH.M-sch=MUON-SACH.M-sch) X [MUON-SACH.M-sch] [ SACH.M-sch, Tn-sch] X SACH [MUON-SACH.M-sch, MUON-SACH.S-th] : (Ngy-mn <27/03/1999) [ Doc-Gia.S-th] MUON-SACH Doc-Gia Hnh 6.3 Cy kt qu biu din vic phn nhm cc biu thc Thut ton ti u ho cu hi trong ngn ng i s quan h V d trn cho ta mt minh ha v vic chuyn i mt cu hi bng ngn ng i s quan h v dng tng ng tt hn (hay ti u hn). Phng php trn tp trung ch yu vo cc php chiu, php chn v tch cc, vi mc ch lm sao y c php chn v php chiu xung mc thp nht, tc l thi hnh cc php ton ny cng sm cng tt, nu c th. Tip theo, kt hp cc php chn vi tch cc thnh php kt ni t nhin lm gim cc kt qu trung gian. Ct li ca vn ti u ho chnh l vic lm gim thiu lu tr trung gian v t lm tng nhanh tc x l cu hi. Tuy nhin, thc hin c cc qu trnh ti u ho nh trn, chng ta cn lu ti th t thc hin cc php ton c th y cc php ton xung cc mc hp l cn thit. Bng di y cho php chng ta cch thc hin cc php bin i tng ng i vi cc php hi (Union), Tr (Minus), Giao (Intersect), Tch cc (Cartesian), Chia(Division), chiu (Projection) v chn (Selection) (B1). Kt ni t nhin tng ng vi dy php tch cc, php chn v php chiu:  EMBED Equation.3  (B2).Php ( kt tng ng vi dy php tch cc v php chn vi iu kin (: Q1(A,B) Q2(C,D) a" (Q1 x Q2): (B(D) (B3).Php giao tng ng vi phn b ca hi hai phn b ca hai quan h:  EMBED Equation.3  v (B4) (B4).Php b ca mt quan h tng ng vi tch cc ca cc php chiu trn tng thuc tnh ca quan h tr i cc b gi tr c trong th hin ca quan h:  EMBED Equation.3  (B5).Thng ca hai quan h tng ng vi hiu ca cc quan h trung gian sau:  EMBED Equation.3  p dng cc cch bin i tng ng trn, kt hp vi cc quy tc y v kt hp nh trnh by trong mc 1.2, chng ta a ra thut ton tng qut ti u ho cc cu hi trong ngn ng i s quan h . Thut ton: u vo (Input): S c php cu hi bng ngn ng i s quan h. u ra (Output): S c php ti u. Bc 1: p dng cc php bin i tng ng nu trong bng (B1) n (B5) trn. Bc 2: p dng lut ( bin i dy cc php chn tng ng: tch php chn thnh cc php chn con. Bc 3: i vi mi php chn, p dng cc lut (,(,( v( nhm y cc php ton chn xung cng su cng tt. Bc 4: i vi mi php chiu, p dng cc quy tc (,( v ( nhm y cc php ton chiu xung cng su cng tt. Bc 5: Tp trung cc php chn nhm p dng lut ( p dng lut ( loi b bt cc php chiu v ch Tp trung cc php chn vi tch cc, nu c, chuyn thnh php kt ni t nhin hay ( kt ni bng cch p dng cc lut ( v (. Nhn xt: 1.Thut gii nu trn ch yu nhm gim khi lng d liu trung gian ch khng ch ra th t thc hin cc php kt ni. v d:  EMBED Equation.3  2.Thut giI ny khng cho chng ta mt kt qu ti u m n ch a ra mt gii php tt. 3.Cc php bin i ch da trn cc php ton c bn l Hi (Union), Tr (Minus), Giao (Intersect), Tch cc (Cartesian), Chia (Division), Chiu (Projection) v Chn (Selection) m chng ta cn c th thc hin cc php bin i da trn cc php ton khc na. Chng VII. C s d liu phn tn 1. Tng quan Nhng nm ca thp k 70, my tnh c kh nng xy dng h thng thng tin v h CSDL. Mt mt hnh thnh v pht trin cc m hnh l thuyt cho h CSDL v mt khc nhng ngun pht trin h thng ng dng ngy cng c nhiu kinh nghim. H thng thng tin hnh thnh trn c s kt ni cc my tnh khc nhau. Nhng nm gn y, h CSDL phn tn c pht trin da trn CSDL v mng my tnh. CSDL phn tn gm nhiu CSDL tch hp li vi nhau thng qua mng my tnh trao i d liu, thng tin CSDL c t chc v lu tr nhng v tr khc nhau trong mng my tnh v chng trnh ng dng lm vic trn c s truy cp d liu nhng im khc nhau . Vn hon ton mi l xy dng v ci t mt CSDL phn tn. V yu cu ca cng ty, doanh nghip, n v kinh doanh v vn t chc sao cho kinh doanh c hiu qu nht v nm bt thng tin nhanh nht khi cc c s ca cng ty hin nhng a im xa nhau cho nn xy dng mt h thng lm vic trn CSDL phn tn l ph hpp xu hng hin nay v h thng ny tho mn c nhng yu cu t chc ca n v. Tc dng ca t chc v k thut ca xu hng pht trin CSDL phn tn l: gii quyt c nhng hn ch ca CSDL tp trung v ph hp xu hng pht trin t nhin vi c cu khng tp trung ca cc t chc, cng ty, doanh nghip Ni mt cch n gin, CSDL phn tn l tp hp d liu logic thuc v cng mt h thng nhng tri rng ra nhiu im trn mng my tnh. Nh vy c hai vn ca CSDL phn tn vi tm quan trng tng ng nhau: Vic phn tn: Trong thc t d liu khng t tn trn cng mt v tr v vy y l c im phn bit CSDL phn tn vi CSDL tp trung v CSDL n l. Lin quan Logic: Trong CSDL phn tn, d liu c mt s c tnh lin kt cht ch vi nhau nh tnh kt ni, tnh lin quan logic Trong CSDL tp trung, mi v tr qun l mt CSDL v ngi s dng phi truy cp n CSDL nhng v tr khc nhau ly thng tin tng hp. Tc dng ca CSDL phn tn: + V t chc v tnh kinh t: T chc phn tn nhiu chi nhnh v dng CSDL phn tn ph hp vi cc t chc kiu ny. Vi vai tr l ng lc thc y kinh t thng mi pht trin rng hn, th vic pht trin cc trung tm my tnh phn tn nhiu v tr tr thnh nhu cu cn thit. + Tn dng nhng CSDL sn c: Hnh thnh CSDL phn tn t cc CSDL tp trung c sn cc v tr. + Thun li cho nhu cu pht trin: gim xung t v chc nng gia cc n v tn ti v gim c xung t gia cc chng trnh ng dng khi truy cp n CSDL. Vi hng tp trung ho, nhu cu pht trin trong tng lai s gp kh khn. + Gim chi ph truyn thng: Trong CSDL phn tn chng trnh ng dng t a phng c th gim bt c chi ph truyn thng khi thc hin bng cch khai thc CSDL ti ch. + Tng s cng vic thc hin: H CSDL phn tn c th tng s lng cng vic thc hin qua p dng nguyn l x l song song vi h thng x l a nhim. Tuy nhin CSDL phn tn cng c tin li trong vic phn tn d liu nh to ra cc chng trnh ng dng ph thuc vo tiu chun m rng v tr lm cho cc ni x l c th h tr nhau. Do trnh c hin tng tc nghn c chai trong mng truyn thng hoc trong cc dch v thng thng ca ton b h thng. + Tnh d hiu v sn sng: Hng pht trin CSDL phn tn cng nhm t c tnh d hiu v tnh sn sng cao hn. Tuy nhin c th t c mc tiu ny khng phi l d lm v i hi s dng k thut phc tp. Kh nng x l t tr ca cc im lm vic khc nhau khng m bo tnh d s dng. 2. M hnh tng qut v CSDL phn tn Trong h thng CSDL phn tn, CSDL cha nhiu my tnh. Cc my tnh trong h thng (mng) lin h vi nhau qua cc phng tin truyn thng. Mi v tr trong h thng c gi l mt trm hoc l nt. Mi nt c th c nhiu my tnh vi kh nng v dung lng khc nhau. H thng cc my tnh ni vi nhau to thnh mng my tnh. Trong mng my tnh, cc my tnh trn mi trm c th khai thc d liu trn cc trm khc v ng thi mi trm c chc nng ring, chu trch nhim bo tr mt h thng CSDL cc b. 3. Thit k mt h Qun tr CSDL phn tn. 3.1. Mc tiu - Tnh cc b x l: l t d liu cng gn cc ng dng s dng cc d liu cng tt. - Mt quan h khng l mt n v phn tn. - Kh nng lu tr c sn ti mi ni. - Chi ph lu tr d liu khng ng k so vi cc chi ph nhp/xut v truyn thng ca cc ng dng. 3.2. Chin lc 3.2.1. To bn sao d liu Ta coi CSDL l mt quan h R th quan h R c lu gi v sao (Copy) trn mi nt ca mng. (D liu c Copy ra lm nhiu bn, c lu tr nhiu ni). 3.2.2. Trong sut v to bn sao Cc bn Copy ca d liu c th c lu nhiu ni nng cao tc thc hin, tin cy. Ngi dng khng h bit v s hin hu ca nhiu bn copy ny. 3.2.3. Phn hoch CSDL 3.2.4. 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1j̲h,h0CJEHOJQJUaJmH sH c=>>> >>> >">(>>>>>>>>>>?%@&@9@ǯǯǡqbMqAqhT+CJOJQJaJ)jh2ih2iCJEHOJQJUaJjwJaM h2iCJUVaJjh2iCJOJQJUaJh2iCJOJQJaJh!.DCJOJQJaJh4CJaJ jqh4h4CJaJ.h4B*CJH*OJQJ^JaJmH phsH +h4B*CJOJQJ^JaJmH phsH 1hwXh4B*CJOJQJ^JaJmH phsH h4aJmH sH 9@:@;@<@=@@@@@@@@@6AyAzAAAAAA˿˿˿tbtRC4h1haO CJOJQJaJhaO haO CJOJQJaJh,3haO 5CJOJQJaJ"h,3haO 56CJOJQJaJhaO CJOJQJaJh1h1CJOJQJaJ)jh2ihCJEHOJQJUaJjTLaM hCJUVaJh1CJOJQJaJh2iCJOJQJaJjh2iCJOJQJUaJ)jh2ih2iCJEHOJQJUaJj.KaM h2iCJUVaJACBJBKBLBXBYBBBBBBBBBBBCC%CRCSCTCUCYCZC~n~_~P~A jh<16CJOJQJaJ jh<16CJOJQJaJ jh<16CJOJQJaJh,3h<165CJOJQJaJh<16CJOJQJaJ jhaO CJOJQJaJ jhaO CJOJQJaJ jhaO CJOJQJaJ jhaO CJOJQJaJ jhaO CJOJQJaJhaO haO CJOJQJaJh,3haO 5CJOJQJaJhaO CJOJQJaJCCCCDD EE,EEFFFFFHfIIKLaMsNNO $$`a$gdvQgdvQ$a$gdFU$7$8$H$`a$gd!.D$7$8$H$`a$gdaO ZCCCCCCCCCCYDZD[DDDDDDDDDD EEE'Evj^^N^jhCJOJQJUaJhCJOJQJaJh2iCJOJQJaJh,3h<165CJOJQJaJ jh<16CJOJQJaJh<16CJaJ jqhh<16CJaJ jh<16CJOJQJaJhaO h<16CJOJQJaJ jh<16CJOJQJaJh,3CJOJQJaJh,3h,35CJOJQJaJh<16CJOJQJaJ'E(E)E*E,EFFFFFFFFFF|G}GGHhH|H}HdI˿zrgYHYHY:H:h<5CJOJQJ^JaJ hhvQCJOJQJ^JaJhCJOJQJ^JaJhhvQOJQJhOJQJhFUhvQOJQJhFUhOJQJh(X=OJQJh!.Dh!.DCJOJQJaJh3CJOJQJaJh !4CJOJQJaJh<16CJOJQJaJjhCJOJQJUaJ)jhhCJEHOJQJUaJj7MaM hCJUVaJdIfIyI}III:J=JzJ}JJJKyKKKKLLLL_MaMMMMpNsNNNNNOOOOOPPQQQQQRRSSSSSS^T`TbTgTŷ﷩者h "CJOJQJ^JaJheeCJOJQJ^JaJh TCJOJQJ^JaJh>>CJOJQJ^JaJhCJOJQJ^JaJhzCJOJQJ^JaJh<5CJOJQJ^JaJ hhvQCJOJQJ^JaJ7OPQQSTTVW!WzWWW4XDX_XXYYYYttt*u$$dh`a$gdvQgdvQ $`a$gdvQgdvQgdvQ $$`a$gdvQgTpTuTTTTTTTTTTTTUUUU~VVVVVVVVVVWWWW!WvWzWWWWWWW0X4XBXDX]X_XXXYYNY^Y_YYYYָָָ֬htCJOJQJ^JaJhhvQCJOJQJaJhXCJOJQJaJhXOJQJhXCJOJQJ^JaJh "OJQJhhvQOJQJ hhvQCJOJQJ^JaJh "CJOJQJ^JaJ8YYYYYYZt]t_trttttt(u*u[u]umuuuuuuuuuu v"v$v5v7vVv]vivpv}vvvvvvvܽyojhFUUaJhh CJOJQJaJ-hh 5CJOJQJ\aJmHnHuh= *h= *CJOJQJaJh= *hvQCJOJQJaJh= *hzcCJOJQJaJUhtCJOJQJ^JaJhhvQOJQJhtOJQJ hhvQCJOJQJ^JaJ*a mt quan h thnh cc tp con ca cc hng trong quan h ban u, mi tp con c mt ngha v mt logic no . VD: Phn tch quan h DU_AN vi MaDV =1, MaDV = 4, MaDV=5. - Phn tch dc: Chia quan h dc theo cc thuc tnh v i khi mt site khng cn tt c cc thuc tnh ca mt quan h. V d: quan h Nhanvien c chia thnh 2 quan h: thng tin c nhn (Ten, Ngaysinh, Diachi, Gioitinh), thng tin v cng vic (MaNV, Luong). 4. X l truy vn phn tn 4.1. Ton vn d liu trong CSDL phn tn 4.2. Giao thc kt thc hai giai on 4.3. Kho phn tn 4.4. Kho hai giai on phn tn 4.5. nh du thi gian 4.6. Khi phc CSDL  TOC \o "1-3" \h \z \u  HYPERLINK \l "_Toc224628959" Chng I: Tng quan v c s d liu  PAGEREF _Toc224628959 \h 1  HYPERLINK \l "_Toc224628960" 1. Mt s khi nim  PAGEREF _Toc224628960 \h 1  HYPERLINK \l "_Toc224628961" 1.1. C s d liu  PAGEREF _Toc224628961 \h 1  HYPERLINK \l "_Toc224628962" 1.2 H qun tr c s d liu (Database Management System- HQTCSDL)  PAGEREF _Toc224628962 \h 1  HYPERLINK \l "_Toc224628963" 2. Cc m hnh d liu  PAGEREF _Toc224628963 \h 2  HYPERLINK \l "_Toc224628964" 2.1. M hnh mng  PAGEREF _Toc224628964 \h 2  HYPERLINK \l "_Toc224628965" 2.2. M hnh phn cp  PAGEREF _Toc224628965 \h 3  HYPERLINK \l "_Toc224628966" 2.3. M hnh quan h :  PAGEREF _Toc224628966 \h 3  HYPERLINK \l "_Toc224628967" 2.4. M hnh thc th lin kt :  PAGEREF _Toc224628967 \h 3  HYPERLINK \l "_Toc224628975" 2.5. M hnh hng i tng :  PAGEREF _Toc224628975 \h 9  HYPERLINK \l "_Toc224628978" Chng II: M hnh c s d liu quan h  PAGEREF _Toc224628978 \h 10  HYPERLINK \l "_Toc224628979" 1.Cc khi nim c bn  PAGEREF _Toc224628979 \h 10  HYPERLINK \l "_Toc224628980" 1.1.Thuc tnh(Attribute):  PAGEREF _Toc224628980 \h 10  HYPERLINK \l "_Toc224628981" 1.2. Quan h  PAGEREF _Toc224628981 \h 11  HYPERLINK \l "_Toc224628982" 1.3. B gi tr (Tuple)  PAGEREF _Toc224628982 \h 12  HYPERLINK \l "_Toc224628983" 1.4.Lc quan h  PAGEREF _Toc224628983 \h 12  HYPERLINK \l "_Toc224628984" 1.5. Th hin ca quan h  PAGEREF _Toc224628984 \h 13  HYPERLINK \l "_Toc224628985" 1.6. Kho - Siu kho - Kho ch nh kho chnh kho ngoi  PAGEREF _Toc224628985 \h 13  HYPERLINK \l "_Toc224628986" 1.7.Ph thuc hm  PAGEREF _Toc224628986 \h 15  HYPERLINK \l "_Toc224628987" 1.8.Rng buc ton vn  PAGEREF _Toc224628987 \h 15  HYPERLINK \l "_Toc224628988" 1.9.Cc thao tc c bn trn cc quan h  PAGEREF _Toc224628988 \h 16  HYPERLINK \l "_Toc224628989" 2.Cc php ton trn i s tp hp  PAGEREF _Toc224628989 \h 17  HYPERLINK \l "_Toc224628990" 2.1.Php hp  PAGEREF _Toc224628990 \h 17  HYPERLINK \l "_Toc224628991" 2.2. Php giao  PAGEREF _Toc224628991 \h 18  HYPERLINK \l "_Toc224628992" 2.3.Php tr  PAGEREF _Toc224628992 \h 18  HYPERLINK \l "_Toc224628993" 2.4.Php tch cc (Descartes)  PAGEREF _Toc224628993 \h 19  HYPERLINK \l "_Toc224628994" 2.5.Php chia  PAGEREF _Toc224628994 \h 20  HYPERLINK \l "_Toc224628995" 3.Cc php ton trn i s quan h  PAGEREF _Toc224628995 \h 21  HYPERLINK \l "_Toc224628996" 3.1.Php chiu (( Projection )  PAGEREF _Toc224628996 \h 21  HYPERLINK \l "_Toc224628997" 3.2. Php chn ( Selection )  PAGEREF _Toc224628997 \h 21  HYPERLINK \l "_Toc224628998" 3.3.Php kt ni ( Join )  PAGEREF _Toc224628998 \h 22  HYPERLINK \l "_Toc224628999" 3.4. Cc php ton kt ni khc  PAGEREF _Toc224628999 \h 23  HYPERLINK \l "_Toc224629003" Chng III  PAGEREF _Toc224629003 \h 33  HYPERLINK \l "_Toc224629004" Ngn ng d liu SQL  PAGEREF _Toc224629004 \h 33  HYPERLINK \l "_Toc224629005" 1. Khi qut v ngn ng d liu SQL  PAGEREF _Toc224629005 \h 33  HYPERLINK \l "_Toc224629006" 2. Cc lnh lin quan n cu trc ca c s d liu  PAGEREF _Toc224629006 \h 33  HYPERLINK \l "_Toc224629007" 2.1 To bng  PAGEREF _Toc224629007 \h 33  HYPERLINK \l "_Toc224629008" 2.2.Xo bng  PAGEREF _Toc224629008 \h 36  HYPERLINK \l "_Toc224629009" 2.3. Thm, xo cc ct ca bng  PAGEREF _Toc224629009 \h 36  HYPERLINK \l "_Toc224629010" 3.Cc lnh cp nht c s d liu  PAGEREF _Toc224629010 \h 36  HYPERLINK \l "_Toc224629011" 3.1.Thm b vo bng  PAGEREF _Toc224629011 \h 36  HYPERLINK \l "_Toc224629012" 3.2.Cp nht ni dung ca b trong bng  PAGEREF _Toc224629012 \h 37  HYPERLINK \l "_Toc224629013" 3.3. Xo cc b trong bng  PAGEREF _Toc224629013 \h 38  HYPERLINK \l "_Toc224629014" 4. Cc lnh truy vn c s d liu  PAGEREF _Toc224629014 \h 39  HYPERLINK \l "_Toc224629015" 4.1. Tm thng tin t cc ct ca bng (php chiu)  PAGEREF _Toc224629015 \h 40  HYPERLINK \l "_Toc224629016" 4.2.Chn cc b ca bng Mnh WHERE (php chn)  PAGEREF _Toc224629016 \h 41  HYPERLINK \l "_Toc224629017" 4.3. Th t hin th cc bn ghi - Mnh ORDER BY  PAGEREF _Toc224629017 \h 42  HYPERLINK \l "_Toc224629018" 4.4. Phn nhm d liu Mnh GROUP BY  PAGEREF _Toc224629018 \h 43  HYPERLINK \l "_Toc224629019" 4.5.iu kin hin th cc bn ghi - Mnh HAVING  PAGEREF _Toc224629019 \h 44  HYPERLINK \l "_Toc224629020" 4.6. Truy vn thng tin t nhiu bng d liu(php kt ni).  PAGEREF _Toc224629020 \h 44  HYPERLINK \l "_Toc224629021" 4.7. Truy vn lng nhau  PAGEREF _Toc224629021 \h 45  HYPERLINK \l "_Toc224629022" 4.8.Cc hm tnh ton trn nhm cc bn ghi  PAGEREF _Toc224629022 \h 49  HYPERLINK \l "_Toc224629023" 4.9. Cc hm tnh ton trn bn ghi  PAGEREF _Toc224629023 \h 50  HYPERLINK \l "_Toc224629025" Chng IV: Rng buc ton vn v ph thuc hm  PAGEREF _Toc224629025 \h 55  HYPERLINK \l "_Toc224629026" 1.Cc vn lin quan n rng buc ton vn  PAGEREF _Toc224629026 \h 55  HYPERLINK \l "_Toc224629027" 1.1. nh ngha  PAGEREF _Toc224629027 \h 55  HYPERLINK \l "_Toc224629028" 1.2.iu kin  PAGEREF _Toc224629028 \h 56  HYPERLINK \l "_Toc224629029" 1.3..Bi cnh  PAGEREF _Toc224629029 \h 57  HYPERLINK \l "_Toc224629030" 1.4.Bng tm nh hng  PAGEREF _Toc224629030 \h 57  HYPERLINK \l "_Toc224629031" 1.5.Hnh ng cn phi c khi pht hin c RBTV b vi phm:  PAGEREF _Toc224629031 \h 58  HYPERLINK \l "_Toc224629032" 2. Cc loi rng buc ton vn  PAGEREF _Toc224629032 \h 59  HYPERLINK \l "_Toc224629033" 2.1. Rng buc ton vn v min gi tr  PAGEREF _Toc224629033 \h 59  HYPERLINK \l "_Toc224629034" 2.2. Rng buc ton vn lin thuc tnh  PAGEREF _Toc224629034 \h 59  HYPERLINK \l "_Toc224629035" 2.3. Rng buc ton vn lin b lin thuc tnh  PAGEREF _Toc224629035 \h 60  HYPERLINK \l "_Toc224629036" 2.4. Rng buc ton vn v ph thuc tn ti  PAGEREF _Toc224629036 \h 60  HYPERLINK \l "_Toc224629037" 2.5. Rng buc ton vn tng hp (lin b - lin quan h)  PAGEREF _Toc224629037 \h 61  HYPERLINK \l "_Toc224629038" 3.ph thuc hm  PAGEREF _Toc224629038 \h 61  HYPERLINK \l "_Toc224629040" 3.1.nh ngha v biu din ph thuc hm  PAGEREF _Toc224629040 \h 61  HYPERLINK \l "_Toc224629041" 3.2.Bao ng ca tp ph thuc hm v h lut dn Armstrong  PAGEREF _Toc224629041 \h 62  HYPERLINK \l "_Toc224629042" 3.3.Bao ng ca tp thuc tnh  PAGEREF _Toc224629042 \h 65  HYPERLINK \l "_Toc224629043" 3.4.Ph v tng ng  PAGEREF _Toc224629043 \h 69  HYPERLINK \l "_Toc224629044" 3.5.Thut ton xc nh kho ca lc quan h  PAGEREF _Toc224629044 \h 73  HYPERLINK \l "_Toc224629046" Chng v: dng chun v cc vn chun ho lc c s d liu quan h  PAGEREF _Toc224629046 \h 78  HYPERLINK \l "_Toc224629047" 1.Dng chun  PAGEREF _Toc224629047 \h 78  HYPERLINK \l "_Toc224629048" 1.1.Thit k km gy nguy him cho CSDL  PAGEREF _Toc224629048 \h 78  HYPERLINK \l "_Toc224629049" 1.2. Phn r  PAGEREF _Toc224629049 \h 80  HYPERLINK \l "_Toc224629050" 1.3.Cc dng chun  PAGEREF _Toc224629050 \h 89  HYPERLINK \l "_Toc224629051" 1.3.2. Cc dng chun  PAGEREF _Toc224629051 \h 90  HYPERLINK \l "_Toc224629052" 2.Chun ho lc c s d liu  PAGEREF _Toc224629052 \h 93  HYPERLINK \l "_Toc224629053" 2.1. Phng php phn r  PAGEREF _Toc224629053 \h 93  HYPERLINK \l "_Toc224629054" 2.2.Phng php tng hp  PAGEREF _Toc224629054 \h 97  HYPERLINK \l "_Toc224629056" Chng VI Ti u ho cu hi  PAGEREF _Toc224629056 \h 104  HYPERLINK \l "_Toc224629057" 1. Cc nguyn tc tng qut ti u ho cu hi:  PAGEREF _Toc224629057 \h 104  HYPERLINK \l "_Toc224629058" 1.1. Cc nguyn tc tng qut  PAGEREF _Toc224629058 \h 104  HYPERLINK \l "_Toc224629059" 1.2.Biu thc tng v cc quy tc  PAGEREF _Toc224629059 \h 104  HYPERLINK \l "_Toc224629060" 2.V d v mt thut ton ti u ho biu thc quan h  PAGEREF _Toc224629060 \h 107  HYPERLINK \l "_Toc224629061" Chng VII. C s d liu phn tn  PAGEREF _Toc224629061 \h 113  HYPERLINK \l "_Toc224629062" 1. Tng quan  PAGEREF _Toc224629062 \h 113  HYPERLINK \l "_Toc224629063" 2. M hnh tng qut v CSDL phn tn  PAGEREF _Toc224629063 \h 115  HYPERLINK \l "_Toc224629064" 3. Thit k mt h Qun tr CSDL phn tn.  PAGEREF _Toc224629064 \h 115  HYPERLINK \l "_Toc224629065" 3.1. Mc tiu  PAGEREF _Toc224629065 \h 115  HYPERLINK \l "_Toc224629066" 3.2. Chin lc  PAGEREF _Toc224629066 \h 115  HYPERLINK \l "_Toc224629067" 4. X l truy vn phn tn  PAGEREF _Toc224629067 \h 116  HYPERLINK \l "_Toc224629068" 4.1. Ton vn d liu trong CSDL phn tn  PAGEREF _Toc224629068 \h 116  HYPERLINK \l "_Toc224629069" 4.2. Giao thc kt thc hai giai on  PAGEREF _Toc224629069 \h 116  HYPERLINK \l "_Toc224629070" 4.3. Kho phn tn  PAGEREF _Toc224629070 \h 116  HYPERLINK \l "_Toc224629071" 4.4. Kho hai giai on phn tn  PAGEREF _Toc224629071 \h 116  HYPERLINK \l "_Toc224629072" 4.5. nh du thi gian  PAGEREF _Toc224629072 \h 116  HYPERLINK \l "_Toc224629073" 4.6. Khi phc CSDL  PAGEREF _Toc224629073 \h 116      PAGE  PAGE 119 MANAGESS Location DEPTS ENO Salary EMPS ENO ENAME DNO DNAME ASSIGNED_TO Hnh 2.1: V d s thc th - lin kt. CEMBED Equation.3T CEMBED Equation.3T THEMBED Equation.3R CSEMBED Equation.3G HSEMBED Equation.3R HREMBED Equation.3C R(CTHRSG) Kho = HS CSEMBED Equation.3G HSEMBED Equation.3C CHEMBED Equation.3R , HREMBED Equation.3C R222(CHS) Kho = SH R221(CHR) Kho = CH,HR CH EMBED Equation.3R HS EMBED Equation.3R HREMBED Equation.3C R22(CHRS) Kho = HS R21(CT) Kho = C THEMBED Equation.3R HSEMBED Equation.3R CEMBED Equation.3T HREMBED Equation.3C R2(CTHRS) Kho = HS R1(CSG) Kho = CS R (( S = R.A=S.A R(ABC) ( S(CDE) = R (( S = B(C *uuuu$v7vXvpvvvvw^ww9xxx>yyy' ! gdYK' ! & ! % ! $d8xa$gd $$`a$gd$7gdvQgdvQ$$dh`a$gdvQvvvvvvvvvvvvwwwwwwww w%w&w'w(wϬݚodJ2jhFUhFU>*B*UmHnHphuh(X=mHnHu jhFUUmHnHujhFUUmHnHuhFUmHnHu#hcN<hFU0JOJQJmHnHu2j hFUhFU>*B*UmHnHphuhFUmHnHuhcN<hFU0JmHnHu$jhcN<hFU0JUmHnHujhFUUaJ hFUaJ(wwXwYwZw[w\w]w^w_w`w|w}w~wwwwwwwwwwwwwwwwӷӤs_NӷӤ jzhFUUmHnHu&hcN<hFU0J6OJQJmHnHu2jhFUhFU>*B*UmHnHphuhcN<hFU0JmHnHuhFUmHnHu$jhcN<hFU0JUmHnHuh(X=mHnHu jhFUUmHnHujhFUUmHnHuhFUmHnHu#hcN<hFU0JOJQJmHnHuwwwxxx3x4x5x6x7x8x9x:x;xWxXxYxZxpxqxrxxӿӀrrXF#hcN<hFU0JOJQJmHnHu2jhFUhFU>*B*UmHnHphuhcN<hFU0JmHnHuhFUmHnHuh(X=mHnHu jthFUUmHnHujhFUUmHnHuhFUmHnHu&hcN<hFU0J6OJQJmHnHu$jhcN<hFU0JUmHnHu2jhFUhFU>*B*UmHnHphuxxxxxxxxxxxxxxxxxxxxxxxxxyyy¹«}rra¹«G2jhFUhFU>*B*UmHnHphu jhhFUUmHnHuhFUmHnHu&hcN<hFU0J6OJQJmHnHu2jhFUhFU>*B*UmHnHphuhcN<hFU0JmHnHuhFUmHnHu$jhcN<hFU0JUmHnHuh(X=mHnHujhFUUmHnHu jnhFUUmHnHuyyyyy8y9y:y;yy?y@y\y]y^y_yuyvywyyyyyyyyyyyyοή팚rοa팚 j\hFUUmHnHu2jhFUhFU>*B*UmHnHphuhcN<hFU0JmHnHuhFUmHnHuh(X=mHnHu jbhFUUmHnHujhFUUmHnHuhFUmHnHu&hcN<hFU0J6OJQJmHnHu$jhcN<hFU0JUmHnHuyyyyyyyyyyyyyyyzzzz9z:z;zUzӿӀrrXB*hcN<hFU0J6OJQJmHnHsH u2jhFUhFU>*B*UmHnHphuhcN<hFU0JmHnHuhFUmHnHuh(X=mHnHu jVhFUUmHnHujhFUUmHnHuhFUmHnHu&hcN<hFU0J6OJQJmHnHu$jhcN<hFU0JUmHnHu2jhFUhFU>*B*UmHnHphuUzVzWzXzYzZz[z\z]zyzzz{z|zzzzzzzzzzzzzzzz¹«‚peeT¹« jJhFUUmHnHuhFUmHnHu#hcN<hFU0JOJQJmHnHuhYK0JOJQJmHnHu2jhFUhFU>*B*UmHnHphuhcN<hFU0JmHnHuhFUmHnHu$jhcN<hFU0JUmHnHuh(X=mHnHujhFUUmHnHu jPhFUUmHnHuy[zz!{{{*|||a}}~|~~39Ss' ! & ! % ! 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