co so du lieu SQL

7/28/2019 co so du lieu SQL

1/228

Khoa HTTT-i hc CNTT 1

i hc Cng Ngh Thng Tin

Mn: C S D LiuThi lng: 45LT+30TH


2/228


Ni dung

Bi 1: Cc khi nim ca mt h CSDL Bi 2: Cc m hnh CSDL

Bi 3: M hnh d liu quan h (ca Codd) Bi 4: Ngn ng i s quan h Bi 5: Ngn ng SQL

Bi 6: Ngn ng tn t Bi 7: Rng buc ton vn trong mt CSDL Bi 8: Ti u ha cu hi bng i s quan h


3/228


Bi 1: Cc khi nim ca mt hCSDL


4/228


Bi 1: Cc khi nim ca mt

h CSDL1. Gii thiu

2. H thng tp tin (File System)3. nh ngha mt CSDL

4. Cc i tng s dng CSDL

5. H qun tr CSDL6. Cc mc ca mt CSDL


5/228


1. Gii thiu

Ti sao cn phi c mt c s d liu?


6/228


2. H thng tp tin

L tp hp cc tp tin ring l phc v cho mt mcch ca n v s dng.

u im: Trin khai ng dng nhanh Kh nng p ng nhanh chng, kp thi (v ch phc v

cho mc ch hn hp)

Khuyt im: Trng lp d liu lng ph, d liu khng nht qun Chi ph cao

Chia s d liu km


7/228Khoa HTTT-i hc CNTT 7

3. C s d liu (1)

nh ngha:

C s d liu l mt h thng cc thng tin ccu trc, c lu tr trn cc thit b lu trnhm tha mn yu cu khai thc thng tin ngthi ca nhiu ngi s dng hay nhiu chng

trnh ng dng vi nhng mc ch khc nhau



3. C s d liu (2)

u im:

Gim trng lp thng tin xung mc thp nht,m bo tnh nht qun v ton vn d liu.

m bo d liu c truy xut theo nhiu cchkhc nhau.

Kh nng chia s thng tin cho nhiungi,nhiu ng dng khc nhau.



3. C s d liu (3)

Nhng vn cn gii quyt:

Tnh ch quyn d liu. Tnh bo mt v quyn khai thc thng tin ca

ngi s dng

Tranh chp d liu

m bo d liu khi c s c



4. Cc i tng s dng

Ngi s dng CSDL khng chuyn v lnh vc tinhc v CSDL -> cn cng c h c th khai thcCSDL khi cn.

Chuyn vin tin hc xy dng cc ng dng phcv cho cc mc ch qun l

Qun tr CSDL: t chc CSDL, bo mt, cp quyn,sao lu, phc hi d liu, gii quyt cc tranh chpd liu



5. H qun tr c s d liu (1)

H qun tr c s d liu (DBMS DataBaseManagement System) l h thng cc phn mm htr tch cc cho cc nh phn tch, thit k v khaithc CSDL.

Cc DBMS thng dng: Visual FoxPro, Microsoft

Access, SQL Server, DB2, Oracle hu ht ccDBMS hin nay u da trn m hnh quan h.




Mt DBMS phi c:

Ngn ng giao tip gia ngi s dng v CSDL

T in d liu (Data Dictionary)

C bin php bo mt khi c yu cu

C ch gii quyt tranh chp d liu

C c ch sao lu (backup), phc hi (restore)

m bo tnh c lp gia d liu v chng trnh




Ngn ng giao tip: Ngn ng m t d liu (DDLData Definition Language):

cho php khai bo cu trc CSDL, cc mi lin h ca d liu,cc quy nh, rng buc d liu. Ngn ng thao tc d liu (DMLData Manipulation

Language): cho php thc hin thao tc thm, xa, sa d liu. Ngn ng truy vn c cu trc (SQL Structured Query

Language): cho php ngi khai thc s dng truy vnthng tin cn thit.

Ngn ng qun l d liu (DCL Data Control Language) chophp thay i cu trc bng, khai bo bo mt, cp quyn cho

ngi s dng.



6. Cc mc biu din 1 CSDL

Mc trong: (mc vt l Physical) l mc lu tr CSDL(cn gii quyt vn g? D liu g? Lu tr nh th

no? u? Cn cc ch mc g? Truy xut tun t hayngu nhin. Dnh cho ngi qun tr v ngi s dngchuyn mn.

Mc quan nim: (Conception hay Logical) cn phi lu

tr bao nhiu loi d liu? l d liu g? mi quan h Mc ngoi: ca ngi s dng v cc chng trnh ng

dng



6. Cc mc biu din 1 CSDL

Cu trc

ngoi 1

Cu trcngoi 2

Cu trcngoi n

Chng trnhng dng n

Mcquan

nimhocmclogic

Mc vt l Cu trc vt l

CSDL

Mi trngthc th gii

thc

NSD1

NSD2



Bi 2: Cc m hnh d liu



Ni dung

1. Gii thiu

2. M hnh d liu mng3. M hnh thc th mi kt hp



1. Gii thiu

M hnh d liu l s tru tng ha ca mitrng thc, biu din d liu mc quan nim.Gii thiu mt s m hnh nh:

M hnh d liu mng

M hnh d liu phn cp

M hnh d liu thc th mi kt hp

M hnh d liu quan h

M hnh d liu hng i tng


19/228


2. M hnh d liu mng (1)

M hnh d liu mng (Network Data Model) cngi tt l m hnh mng hoc m hnh li l mhnh c biu din bi mt th c hng. Mu tin (record)

Loi mu tin

Loi lin h (set type) Bn s


20/228



Mu tin: m t 1 i tng trong th gii thc.(NV001,Nguyen Lam,Nam,10/10/1970,Dong Nai)

Loi mu tin: l 1 tp cc mu tin c cng tnhcht. V d: NHANVIEN K hiu:

Loi lin h: m t s lin kt gia 1 loi mu tinch v 1 loi mu tin thnh vin K hiu:

NHANVIEN

Tham gia

CONGVIEC


21/228



Bn s:ch ra s lng cc mu tin tham gia trongmi lin h (1:1) (one-to-one): mi mu tin ca loi mu tin ch kthp vi ng 1 mu tin ca loi mu tin thnh vin. (1:n) (one-to-many): mi mu tin ca loi mu tin ch

kt hp vi 1 hay nhiu mu tin thnh vin.

(n:1) (many-to-one): nhiu mu tin ca loi mu tin chkt hp vi ng 1 mu tin ca loi mu tin thnh vin. (Recursive): mt loi mu tin ch cng c th ng thi

l loi mu tin thnh vin vi chnh n. Loi lin h nyl quy


22/228



PHONG

CONGVIEC

NHANVIEN

LYLICH

gm c

qun ltrc tipcng lm

n:1

1:11:n

1:n


23/228



M hnh d liu mng: Tng i n gin

D s dng

Khng thch hp biu din CSDL c quy m ln

Kh nng din t ng ngha km


24/228



Bi tp:

Xy dng m hnh d liu mng cho c sd liu qun l bn hng trong mt siu th(gio vin m t hin trng, xc nh cc yucu)


25/228


3. M hnh thc th mi kt hp

3.1 Gii thiu3.2 Loi thc th, thc th

3.3 Thuc tnh ca loi thc th3.4 Kho ca loi thc th3.5 Loi mi kt hp, mi kt hp

3.6 Thuc tnh ca loi mi kt hp3.7 Bn s3.8 M hnh ER m rng


26/228


3.1 Gii thiu

M hnh thc th mi kt hp (Entity-Relationship Modelvit tc ER)c CHENgii thiu nm 1976.

M hnh ER c s dng nhiu trong thit

k d liu mc quan nim.


27/228


3.2 Loi thc th

nh ngha: loi thc th (Entity Type) lnhng loi i tng hay s vt ca th gii

thc tn ti c th cn c qun l. V d: HOCVIEN, LOP, MONHOC, K hiu:

HOCVIEN LOP


28/228


29/228


3.3 Thuc tnh ca loi thc th

(Entity Attribute)

nh ngha: thuc tnh l nhng tnh chtc trng ca loi thc th cn lu tr.

V d: Loi thc th HOCVIEN c ccthuc tnh: M hc vin, h tn, gii tnh,ngy sinh, ni sinh

K hiu:HOCVIEN

HotenGioitinh

Mahv

NgaysinhNoisinh


30/228


3.3 Cc loi thuc tnh (1)

n tr (Simple): mi thc th ch c mt gi tr ng vi ccthuc tnh.

V d: Mahv,Hoten a hp (Composite): thuc tnh c th c to thnh t

nhiu thnh phn.V d: DCHI(SONHA,DUONG,PHUONG,QUAN)

hay thuc tnh HOTEN(HO,TENLOT,TEN). a tr (Multi-valued): thuc tnh c th c nhiu gi tr i

vi mt thc th.V d: BANGCAP k hiu {BANGCAP}


31/228


3.3 Cc loi thuc tnh (2)

Tm li, cc thuc tnh a hp v a tr cth lng nhau ty . V d: thuc tnh BANGCAP ca HOCVIEN l

mt thuc tnh a hp c k hiu bng

{BANGCAP(TRUONGCAP,NAM,KETQUA,CHUYENNGANH)}


32/228


3.4. Kho ca loi thc th

(entity type key)

Kha ca loi thc th l thuc tnh nhndin thc th.

Cn c vo gi tr ca kha c th xc nhduy nhtmt thc th.

V d: Mi hc vin c mt m s duy nht => Kho

ca loi thc th HOCVIEN l M hc vin


33/228


34/228


3.5 Loi mi kt hp (2)

Gia hai loi thc th c th tn ti nhiuhn mt loi mi kt hp.

V dThuc

HOCVIEN LOP

L trng lp


35/228


3.5 S ngi ca loi mi kt hp

(relationship degree)

S ngi ca loi mi kt hpl s loi thcth tham gia vo loi mi kt hp .

V d 1: Loi mi kt hp Thuckt hp 2loi thc th HOCVIEN v LOP nn c sngi l 2.

V d 2: Loi mi kt hp Ging dykthp 3 loi thc th GIAOVIEN, MONHOC,LOP nn c s ngi l 3.


36/228


37/228


3.6 Thuc tnh ca loi mi kt hp

(relationship type attribute)

Thuc tnh ca loi mi kt hp bao gm ccthuc tnh kho ca cc loi thc th tham gia vo

loi mi kt hp . Ngoi ra cn c th c thmnhng thuc tnh b sung khc.

V d: Loi mi kt hp Ging dygia ba loi

thc th GIAOVIEN, MONHOC v LOP c ccthuc tnh l Magv,Mamh,Malop, ngoi ra cn cthuc tnh ring l Hocky, Nam


38/228


LOP MONHOC

Giang day

GIAOVIEN

Hocky

Nam


39/228


3.7 Bn s(relationship cardinality)

Loi mi kt hp th hin lin kt gia ccthc th, mi lin kt c gi l mt nhnh.

nh ngha: bn s ca nhnh l s lng tithiu v s lng ti a cc thc th thucnhnh tham gia vo loi mi kt hp.

K hiu: (bn s ti thiu, bn s ti a) V d: Loi thc th NhanVien v DeAn c

loi mi kt hp ThamGia.


40/228


Thuc

HOCVIEN LOP

(1,1) (1,n)


41/228


Thuc

HOCVIEN LOP

L trng lp

(1,1) (1,n)

(0,1) (1,1)


42/228


43/228


3.7.1 Chuyn bit ha (tng

qut ha)

GiaoVien HocVien

ConNguoi

HocVi

NgayVL

NgayNH

Kha

SoCMND

HoTen

...


44/228


45/228


nh ngha: L loi thc th khng c thuc tnh kha

Phi tham gia trong mt loi mi kt hp xc nh trong c mt loi thc th ch.

K hiu:

V d:loi thc th LANTHI c thuc tnh Ln vtham gia trong loi mi kt hp Thivi loi thcth HOCVIEN v MONHOC l loi thc th yu.

Thc th

3.7.3 Loi thc th yu


46/228


3.7.3 Loi thc th yu

Thi

HOCVIEN LANTHI

(1,n) (1,n)

MONHOC

(1,n)


47/228


Bi tp

Xy dng m hnh ER Xy dng m hnh ER cho CSDL qun l

gio v gm c cc chc nng sau: Lu tr thng tin: Hc vin , gio vin, mn hc Xp lp cho hc vin, chn lp trng cho lp Phn cng ging dy: gio vin dy lp no vi

mn hc g, hc k, nm hc no. Lu tr kt qu thi: hc vin thi mn hc no,

ln thi th my, im thi bao nhiu.


48/228


Bi 3: M hnh d liu quan h(Relational Data Model)


49/228


1. Gii thiu

2. Cc khi nim 2.1 Thuc tnh

2.2 Quan h

2.3 B gi tr

2.4 Th hin ca quan h

2.5 Tn t

2.6 Php chiu 2.7 Kha

2.8 Lc quan h v lc CSDL

2.9 Hin thc m hnh ER bng m hnh d liu quan h.

Ni dung


50/228


1. Gii thiu

M hnh D liu Quan h (Relational DataModel) da trn khi nim quan h.

Quan hl khi nim ton hc da trn nntng hnh thc v l thuyt tp hp.

M hnh ny do TS. E. F. Codd a ra nm1970.


51/228


2.1 Thuc tnh (attribute)

Thuc tnh: Tn gi: dy k t (gi nh)

Kiu d liu: S, Chui, Thi gian, Lun l, OLE. Min gi tr: tp gi tr m thuc tnh c th nhn. K

hiu min gi tr ca thuc tnh A l Dom(A).

V d:GIOITINH kiu d liu l Chui,min gi tr

Dom(GIOITINH)=(Nam,Nu) Ti mt thi im, mt thuc tnh khng c gi tr

hoc cha xc nh c gi tr => gi tr Null


52/228


2.2 Quan h (relation)

nh ngha: quan h l mt tp hu hn ccthuc tnh. K hiu: Trong Q l tn quan h, l tp

cc thuc tnh ca quan h Q

V d:HOCVIEN (Mahv, Hoten, Ngsinh, Gioitinh, Noisinh, Malop)LOP (Malop, Tenlop, Siso, Trglop, Khoa)

),...,,( 21 nAAAQ

nAAAQ ,...,, 21


53/228


2.3 B (tuple)

nh ngha:B l cc thng tin ca mt i tngthuc quan h, c gi l mu tin (record), dng.

Quan h l mt bng (table) vi cc ct l ccthuc tnh v mi dng c gi l b.

Mt b ca quan h lvi

V d: HOCVIEN(Mahv, Hoten, Ngsinh, Noisinh)c q=(1003,Nguyen Van Lam, 1/1/1987,Dong Nai)ngha l hc vin c m s l 1003, h tn l

Nguyen Van Lam, sinh ngy 1/1/1987 Dong Nai

),...,,( 21 nAAAQ ),...,,( 21 nAAAQ

)( ii ADoma


54/228


2.4 Th hin ca quan h

(instance) nh ngha: th hin ca mt quan h l tp hp cc

b gi tr ca quan h ti mt thi im nht nh.

K hiu:th hin ca quan h Q lTQ V d:THOCVIENl th hin ca quan h HOCVIEN ti

thi im hin ti gm c cc b nh sau:

HOCVIENMahv HoTen Gioitinh Noisinh Malop

K1103 Ha Duy Lap Nam Nghe An K11

K1102 Tran Ngoc Han Nu Kien Giang K11

K1104 Tran Ngoc Linh Nu Tay Ninh K11


55/228


2.5 Tn t

nh ngha: tn t l mt quy tc dng m t mt quan h.

K hiu: ||Q|| V d: THI (Mahv, Mamh, Lanthi, Diem)

||THI||: mi hc vin c php thi mt mn

hc nhiu ln, mi ln thi lu tr hc vinno thi mn g? ln thi th my? v im lbao nhiu?


56/228


2.6 Php chiu (1)

Php chiu: Dng trch gi tr ca mt s thuc tnhtrong danh sch cc thuc tnh ca quan h.

K hiu: php chiu ca quan h R ln tp thuc tnh Al R[A] hoc R.A.

V d:

hv1=

hv2 =

hv3 =

HOCVIEN

Mahv HoTen Gioitinh Noisinh Malop





57/228


2.6 Php chiu (2)

Php chiu ca quan h HOCVIEN ln thuctnh NoiSinh ca quan h HOCVIEN:

HOCVIEN[Noisinh] = {Nghe An,Kien Giang,Tay Ninh}

HOCVIEN






58/228


Php chiu ln 1 tp thuc tnhK={Hoten,Noisinh} ca quan h HOCVIEN

HOCVIEN[Hoten, Noisinh] = {{Ha Duy Lap, NgheAn},{Tran Ngoc Han, Kien Giang},{Tran NgocLinh,Tay Ninh}}

HOCVIEN





2.6 Php chiu (3)


59/228


60/228


Php chiu 1b ln 1 thuc tnh

hv1[Hoten] = {Ha Duy Lap}

HOCVIEN



K1102 Tran Ngoc Han Nu Kien Giang K11K1104 Tran Ngoc Linh Nu Tay Ninh K11

hv1=

hv2=hv3=

2.6 Php chiu (5)


61/228


2.6 Php chiu (6)

Php chiu 1b ln 1 tp thuc tnh tp thuc tnh K={Hoten, Gioitinh} hv2[K] = {Tran Ngoc Han,Nu}

hv1 = hv2 =

hv3 =

HOCVIEN


K1103 Ha Duy Lap Nam Nghe An K11K1102 Tran Ngoc Han Nu Kien Giang K11



62/228


2.7 Kha

2.7.1 Siu kha (super key)

2.7.2 Kha (key)

2.7.3 Kha chnh (primary key)

2.7.5 Kha ngoi (foreign key)


63/228


2.7.1 Siu kha (super key) (1)

Siu kha: l mt tp con cc thuc tnh ca Q+m gi tr ca chng c th phn bit 2b khc

nhau trong cng mt th hinTQbt k.Ngha l: t1, t2TQ, t1[K] t2[K] K l siukha ca Q.

Mt quan h c t nht mt siu kha (Q+

) v c thc nhiu siu kha.


64/228


2.7.1 Siu kha (super key) (2)

V d: cc siu kha ca quan h HOCVIEN l:{Mahv};{Mahv,Hoten};{Hoten};{Noisinh,Hoten}

HOCVIEN





K1105 Tran Minh Long Nam TpHCM K11

K1106 Le Nhat Minh Nam TpHCM K11


65/228


66/228


V d: cc siu kha ca quan h HOCVIEN l:{Mahv};{Mahv,Hoten};{Hoten};{Hoten,Gioitinh};

{Noisinh,Hoten};{Mahv,Hoten,Gioitinh,Noisinh}

=> th kha ca quan h HOCVIEN c th l{Mahv}; {Hoten}

V d: kha ca quan h GIANGDAY lK={Magv,Mamh,Malop}. Nh vy thuc tnh khas l: Magv,Mamh,Malop.

2.7.2 Kha (key) (2)


67/228


2.7.3 Kha chnh (primary key)

Khi ci t trn mt DBMS c th, nu quan h cnhiu hn mt kha, ta ch c chn mt v gi l

kha chnh K hiu: cc thuc tnh nm trong kha chnh khi

lit k trong quan h phi c gch di.

V d: HOCVIEN (Mahv,Hoten,Gioitinh,Noisinh,Malop)

GIANGDAY(Magv,Mamh,Malop,Hocky,Nam)


68/228


69/228


70/228


2.7.4 Kha ngoi (3)

HOCVIEN







LOP

Malop Tenlop Trglop Siso Magvcn

K11 Lop 1 khoa 1 K1106 11 GV07

K12 Lop 2 khoa 1 K1205 12 GV09

K13 Lop 3 khoa 1 K1305 12 GV14


71/228


72/228


2.8 Lc quan h (2)

Lc quan h c c trng bi: Mt tn phn bit

Mt tp hp hu hn cc thuc tnh (A1, , An)

K hiu ca lc quan h Qgm nthuctnh (A1, A2, ... An) l :

Q(A1, A2, ..., An)


73/228


2.8 Lc quan h (3)

HOCVIEN(Mahv,Hoten,Gioitinh,Noisinh,Malop

Tn t: mi hc vin c mt m hc vin phn bit

vi cc hc vin khc. Cn lu tr h tn, gii tnh, nisinh v thuc lp no.

HOCVIEN


K1103 Ha Duy Lap Nam Nghe An K11K1102 Tran Ngoc Han Nu Kien Giang K11





74/228


75/228


HOCVIEN(MAHV, HO, TEN, NGSINH, GIOITINH, NOISINH, MALOP)

Tn t: mihc vin phn bitvi nhau bng m hc vin, lu trh tn, ngy sinh,gii tnh, ni sinh, thuclp no.

LOP(MALOP, TENLOP, TRGLOP, SISO, MAGVCN)Tn t: milpgm c m lp, tn lp,hc vin lm lptrngcalp,sslp vgio vin chnhim.

KHOA(MAKHOA, TENKHOA, NGTLAP, TRGKHOA)

Tn t: mi khoa cn lu tr m khoa, tn khoa, ngy thnh lp khoa v trng khoa(cng l mt gio vin thuc khoa).

MONHOC(MAMH, TENMH, TCLT, TCTH, MAKHOA)Tn t:mi mn hccnlutr tn mn hc,s tn ch l thuyt,s tn chthc hnh

v khoa noph trch.

DIEUKIEN(MAMH, MAMH_TRUOC)

Tn t: c nhng mn hchc vinphi c kinthctmts mn hctrc.

Lc CSDL qun l sinhvin


76/228


GIAOVIEN(MAGV,HOTEN,HOCVI,HOCHAM,GIOITINH,NGSINH,NGVL,

HESO, MUCLUONG, MAKHOA)

Tn t: m gio vin phnbitgia cc gio vin, cnlutrh tn, hcv,hc

hm, gii tnh, ngy sinh, ngy vo lm, hs,mclng v thucmt khoa.

GIANGDAY(MALOP,MAMH,MAGV,HOCKY, NAM,TUNGAY,DENNGAY)

Tn t: mihckcanmhcs phn cng gingdy: lp no hc mn g dogio vin noph trch.

KETQUATHI(MAHV, MAMH, LANTHI, NGTHI, DIEM, KQUA)

Tn t: lu trktqu thi cahc vin: hc vin no thi mn hc g, ln thi thmy, ngy thi l ngy no, im thi bao nhiu v ktqu l t hay khng t.


77/228


78/228


79/228


80/228


2. Biu thc SQH

Biu thcSQH l mtbiu thcgm cc phpton SQH.

Biu thc SQH c xem nh mt quan h(khng c tn).

C tht tn cho quan hc to tmtbiu

thcSQH. C thi tn cc thuc tnh ca quan hcto

tmtbiuthcSQH.


81/228


3. Cc php ton

3.1 Gii thiu3.2 Php chn

3.3 Php chiu3.4 Php gn

3.5 Cc php ton trn tp hp

3.6 Php kt3.7 Php chia3.8 Hm tnh ton v gom nhm


82/228


3.1 Gii thiu (1)

C nm php ton c bn: Chn ( ) hoc ( : )

Chiu ( ) hoc ( [] ) Tch ( )

Hiu ( )

Hi ( )


83/228


3.1 Gii thiu (2)

Cc php ton khc khng c bn nhng hu ch: Giao ( )

Kt ( ) Chia ( )

Php b ( )

i tn ( )

Php gn ( ) Ktqu sau khi thchin cc php ton l cc quan

h, do c thkthpgia cc php ton tonn php ton mi.


84/228


3.2 Php chn (Selection)

Trch chn cc b (dng) t quan h R. Cc bctrch chnphitha mn iukinchn p.

K hiu: nhngha: p(t):thaiukin p

Ktqutrv l mt quan h, c cng danh sch thuc

tnh vi quan h R. Khng c ktqu trng. Php chn c tnh giao hon

)(Rp

)}(,/{)( tpRttRp

)())(())(()21(1221

RRRpppppp


85/228


Lc CSDL qun l gio v


LOP(MALOP, TENLOP, TRGLOP, SISO, MAGVCN)

KHOA(MAKHOA, TENKHOA, NGTLAP, TRGKHOA)MONHOC(MAMH, TENMH, TCLT, TCTH, MAKHOA)







86/228


3.2 V d php chn

Tm nhnghc vin Nam c ni sinh TpHCM

(Gioitinh=Nam)(Noisinh=TpHCM)(HOCVIEN)

HOCVIEN








87/228


3.3 Php chiu (Project)

S dng trch chn gi tr mt vi thuc tnh caquan h

K hiu:

trong Ail tn cc thuc tnh c chiu. Kt qu tr v mt quan h c k thuc tnh theo th

t nh lit k. Cc dng trng nhau ch ly mt. Php chiu khng c tnh giao hon

)(,...,2

,1

Rk

AAA


88/228


3.3 V d

Tm m s, h tn nhng hc vin Nam c ni sinh TpHCM

Mahv,Hoten(Gioitinh=Nam)(Noisinh=TpHCM)(HOCVIEN)

HOCVIEN








89/228


90/228


91/228


3.5.1 Gii thiu

Cc php ton thc hin trn 2 quan h xut pht t lthuyt tp hp ca ton hc: php hi (RS), phpgiao (RS), php tr (R-S), php tch (RS).

i vi cc php hi, giao, tr, cc quan h R v Sphi kh hp: S lng thuc tnh ca R v S phi bng nhau:

R(A1,A2,An) v S(B1,B2,Bn) Min gi tr ca thuc tnh phi tng thch

dom(Ai)=dom(Bi)

Quan h kt qu ca php hi, giao, tr c cng tnthuc tnh vi quan h u tin.


92/228


3.5.2 Php hi (Union)

}|{ StRttSR

DOT1

Mahv Hoten

K1103 Le Van Tam

K1114 Tran Ngoc Han

K1203 Le Thanh Hau

K1308 Nguyen Gia

DOT2

Mahv Hoten

K1101 Le Kieu My

K1114 Tran Ngoc Han

Mahv Hoten

K1101 Le Kieu My

K1103 Le Van Tam

K1114 Tran Ngoc Han

K1203 Le Thanh Hau

K1308 Nguyen Gia

DOT1DOT2

K hiu: RS

nh ngha: trong R,S l

hai quan h kh hp. V d: Hc vin c khen thng t 1 hoc t 2


93/228


94/228


95/228


96/228


3.5.5 Php tch (2)

HOCVIEN

Mahv Hoten

K1103 Le Van Tam

K1114 Tran Ngoc Han

K1203 Le Thanh Hau

MONHOC

Mamh

CTRR

THDC

CTDL

Mahv Hoten Mamh

K1103 Le Van Tam CTRR

K1114 Tran Ngoc Han CTRR

K1203 Le Thanh Hau CTRR

K1103 Le Van Tam THDC

K1114 Tran Ngoc Han THDC

K1203 Le Thanh Hau THDC

K1103 Le Van Tam CTDL

K1114 Tran Ngoc Han CTDL

K1203 Le Thanh Hau CTDL

HOCVIENMONHOC


97/228


3.6 Php kt

3.6.1 Php kt

3.6.2 Php kt bng, php kt t nhin

3.6.3 Php kt ngoi


98/228


3.6.1 Php kt (Theta-Join) (1)

Theta-join (): Tng t nh php tch kt hp viphp chn. iu kin chn gi l iu kin kt.

K hiu:trong R,S l cc quan h, p l iu kin kt

Cc b c gi tr NULL ti thuc tnh kt ni khngxuthin trong ktquca php kt.

Php kt vi iu kin tng qut gi l -kt vi lmt trong nhng php so snh (,,,,,)

SRp


99/228


100/228


3.6.2 Php kt bng, kt t nhin

Nu l php so snh bng (=), php kt gi lphp kt bng (equi-join).

K hiu: Nu iu kin ca equi-join l cc thuc tnh ging

nhau th gi l php kt t nhin (natural-join). Khi kt qu ca php kt loi b bt 1 ct (b 1trong 2 ct ging nhau)

K hiu: hoc

LOPHOCVIENTrglopMahv

KETQUATHIHOCVIENMahv

KETQUATHIHOCVIENMahv

*


101/228


3.6.3 Php kt ngoi (outer join)

M rng php kt trnh mt thng tin Thc hin php kt v sau thm vo kt qu ca

php kt cc b ca quan h m khng ph hp vicc b trong quan h kia. C 3 loi:

Left outer join R S

Right outer join R S Full outer join R S

V d: In ra danh sch tt c cc hc vin v ims ca cc mn hc m hc vin thi (nu c)


102/228


3.6.3 Php kt ngoi (2)

HOCVIEN KETQUATHImahv

KETQUATHI

Mahv Mamh Diem

HV01 CSDL 7.0

HV02 CSDL 8.5

HV01 CTRR 8.5

HV03 CTRR 9.0

HOCVIEN

Mahv Hoten

HV01 Nguyen Van Lan

HV02 Tran Hong Son

HV03 Nguyen Le

HV04 Le Minh

Mahv Hoten Mahv Mamh Diem

HV01 Nguyen Van Lan HV01 CSDL 7.0

HV01 Nguyen Van Lan HV01 CTRR 8.5

HV02 Tran Hong Son HV02 CSDL 8.5

HV03 Nguyen Le HV03 CTRR 9.0

HV04 Le Minh Null Null Null


103/228


nhngha:

R v S l hai quan h, R+ v S+ ln lt l tp

thuc tnh ca R v S. iukin S+

l tp conkhng bngca R+. Q l ktqu php chia gia Rv S, Q+ = R+ - S+

C thdintbng php ton isnh sau:

3.7 Php chia (Division)

}),(,/{ RstSstSRQ

21

12

1

))((

)(

TTT

RTST

RT

SR

SR


104/228


3.7 Php chia (2)

Mahv

HV01

HV03

KETQUATHI

Mahv Mamh Diem

HV01 CSDL 7.0

HV02 CSDL 8.5

HV01 CTRR 8.5

HV03 CTRR 9.0

HV01 THDC 7.0

HV02 THDC 5.0

HV03 THDC 7.5

HV03 CSDL 6.0

MONHOC

Mamh Tenmh

CSDL Co so du lieu

CTRR Cau truc roi rac

THDC Tin hoc dai cuong

KETQUA

MONHOC

KETQUAMONHOC

][

],[

MamhMONHOCMONHOC

MamhMahvKETQUATHIKETQUA

3 8 Hm tnh ton v gom nhm


105/228


3.8 Hm tnh ton v gom nhm(1)

Hm tnh ton gm cc hm: avg(giatri),min(giatri), max(giatri), sum(giatri), count(giatri).

Php ton gom nhm:

E l biu thc i s quan h

Gil thuc tnh gom nhm (rng, nu khng gom nhm) Fi l hm tnh ton

Ail tn thuc tnh

)()(),...,(),(,...,, 221121 Ennn AFAFAFGGG

3 8 Hm tnh ton v gom nhm


106/228


3.8 Hm tnh ton v gom nhm(2)

im thi cao nht, thp nht, trung bnh camn CSDL ?

im thi cao nht, thp nht, trung bnh ca

tng mn ?)()(),min(),max( KETQUATHIDiemavgDiemDiemMamh

)(CSDL''Mamh)(),min(),max( KETQUATHIDiemagvDiemDiem


107/228

M t cc cu truy vn sau


108/228


M t cc cu truy vn saubng SQH

1. In ra danh sch cc sn phm (MASP,TENSP) do Trung Qucsn xut c gi t 30.000 n 40.000

2. In ra danh sch cc khch hng (MAKH, HOTEN) mua hngtrong ngy 1/1/2007.

3. In ra danh sch cc sn phm (MASP,TENSP) do Trung Quocsn xut hoc cc sn phm c bn ra trong ngy 1/1/2007.

4. Tm cc s ha n mua cng lc 2 sn phm c m s BB01 v

BB02.5. In ra danh sch cc sn phm (MASP,TENSP) do Trung Quoc

sn xut khng bn c trong nm 2006.6. Tm s ha n mua tt c cc sn phm do Singapore sn xut


109/228


110/228


Cu 2

In ra danh sch cc khch hng (MAKH,

HOTEN) mua hng trong ngy 1/1/2007.

],[)#2007/1/1#(: hotenmakhnghdHOADONKHACHHANGMAKH

)()#2007/1/1#(, KHACHHANGHOADONMAKH

nghdhotenmasp


111/228


112/228


Cu 4

Tm cc s ha n mua cng lc cc snphm c m s BB01 v BB02.

BAC

sohdBBmaspCTHDB

sohdBBmaspCTHDA

])['02'(:

])['01'(:

BAC

CTHDB

CTHDA

BBmaspsohd

BBmaspsohd

)(

)(

'02'

'01'

Hoc


113/228

C 6


114/228


Cu 6

ABC

CTHDSANPHAMB

SANPHAMA

MASP

Singaporenuocsxsohdmasp

Singaporenuocsxmasp

)(

)(

'',

''

Tm s ha n mua tt c cc sn phmdo Singapore sn xut


115/228


116/228

1 Gii hi


117/228


L ngn ng chun truy vn v thao tc trnCSDL quan h

L ngn ng phi th tc Khi ngun ca SQL l SEQUEL - Structured

English Query Language, nm 1974) Cc chun SQL

SQL89 SQL92 (SQL2)

SQL99 (SQL3)

1. Gii thiu

2 C i i


118/228


2. Cc ngn ng giao tip

Ngn ng nh ngha d liu (Data DefinitionLanguage - DDL): cho php khai bo cu trc bng,cc mi quan h v cc rng buc.

Ngn ng thao tc d liu (Data ManipulationLanguage - DML): cho php thm, xa, sa d liu.

Ngn ng truy vn d liu (Structured Query

LanguageSQL): cho php truy vn d liu. Ngn ng iu khin d liu (Data Control

LanguageDCL): khai bo bo mt thng tin, cpquyn v thu hi quyn khai thc trn c s d liu.

3 N h h d li


119/228


3.1 Lnh to bng (CREATE)3.1.1 C php

3.1.2 Mt s kiu d liu3.2 Lnh sa cu trc bng (ALTER)

3.2.1 Thm thuc tnh3.2.2 Sa kiu d liu ca thuc tnh

3.2.3 Xo thuc tnh3.2.4 Thm rng buc ton vn3.2.5 Xo rng buc ton vn

3.3 Lnh xa bng (DROP)

3. Ngn ng nh ngha d liu

3 1 L h t b


120/228


3.1.1 C php

CREATE TABLE (

[not null], [not null],

[not null],khai bo kha chnh, kha ngoi, rng buc)

3.1 Lnh to bng


121/228

3 1 L h t b (3)


122/228


3.1 Lnh to bng (3)

Lc CSDL qun l bn hng gm c cc quan h sau:

KHACHHANG (MAKH, HOTEN, DCHI, SODT, NGSINH,DOANHSO, NGDK, CMND)

NHANVIEN (MANV,HOTEN, NGVL, SODT)

SANPHAM (MASP,TENSP, DVT, NUOCSX, GIA)

HOADON (SOHD, NGHD, MAKH, MANV, TRIGIA)

CTHD (SOHD,MASP,SL)

3 1 L h t b (4)


123/228


3.1 Lnh to bng (4)

Create table KHACHHANG

(

MAKH char(4) primary key,

HOTEN varchar(40),

DCHI varchar(50),

SODT varchar(20),

NGSINH smalldatetime,

DOANHSO money,

NGDK smalldatetime,

CMND varchar(10)

)

3 1 L h t b (5)


124/228


3.1 Lnh to bng (5)

Create table CTHD

(

SOHD int foreign keyreferences HOADON(SOHD),

MASP char(4) foreign keyreferences SANPHAM(MASP),

SL int,constraint PK_CTHD primary key (SOHD,MASP)

)

3 2 S t b (1)


125/228


3.2.1 Thm thuc tnhALTER TABLE tnbng ADD tnct kiudliu

V d: thm ct Ghi_chu vo bng khch hngALTER TABLE KHACHHANG ADD GHI_CHU varchar(20)

3.2.2 Sa kiu d liu thuc tnhALTER TABLE tnbng ALTER COLUMN tnct

kiudliu_mi

Lu :

Khng phi sa bt k kiu d liu no cng c

3.2 Sa cu trc bng(1)

3 2 S t b (2)


126/228


V d: Sa Ct Ghi_chu thnh kiu d liu varchar(50)ALTER TABLE KHACHHANG ALTER COLUMN GHI_CHU varchar(50)

Nu sa kiu d liu ca ct Ghi_chu thnh varchar(5), m

trc nhp gi tr cho ct Ghi_chu c di hn 5kt th khng c php.

Hoc sa t kiu chui k t sang kiu s,

3.2.3 Xa thuc tnhALTER TABLE tn_bng DROP COLUMN tn_ct

V d: xa ct Ghi_chu trong bng KHACHHANGALTER TABLE NHANVIEN DROP COLUMN Ghi_chu


3 2 Sa c trc bng(3)


127/228


3.2.4 Thm rng buc ton vn

ALTER TABLE ADD CONSTRAINT

UNIQUE tn_ct

PRIMARY KEY (tn_ct)

FOREIGN KEY (tn_ct)REFERENCES tn_bng(ct_l_kha_chnh) [ONDELETE CASCADE] [ON

UPDATE CASCADE]

CHECK (tn_ctiu_kin)



128/228

3 2 Sa cu trc bng(5)


129/228


3.2.5 Xa rng buc ton vnALTER TABLE tn_bng DROP CONSTRAINT

tn_rng_buc V d:

Alter table CTHD drop constraint FK_CT_SP

Alter table SANPHAM drop constraint ck_gia

Lu : i vi rng buc kha chnh, mun xarng buc ny phi xa ht cc rng buc khangoi tham chiu ti n


3 3 Lnh xa bng


130/228


C php

DROP TABLE tn_bng

V d: xa bng KHACHHANG.DROP TABLE KHACHHANG

Lu : khi mun xa mt bng phi xa tt

c nhng kha ngoi tham chiu ti bng trc.

3.3 Lnh xa bng

4 Ngn ng thao tc d liu


131/228


Gm cc lnh:

4.1 Lnh thm d liu (INSERT)

4.2 Lnh sa d liu (UPDATE)

4.3 Lnh xa d liu (DELETE)

4. Ngn ng thao tc d liu

4 1 Thm d liu


132/228


C php

INSERT INTO tn_bng (ct1,,ctn) VALUES(gi_tr_1,., gi_tr_n)

INSERT INTO tn_bng VALUES (gi_tr_1,gi_tr_2,, gi_tr_n)

V d: insert into SANPHAM values('BC01','But chi', 'cay',

'Singapore', 3000)

insert into SANPHAM(masp,tensp,dvt,nuocsx,gia)

values ('BC01','But chi','cay','Singapore',3000)

4.1 Thm d liu

4 2 Sa d liu


133/228


C php

UPDATE tn_bngSET ct_1 = gi_tr_1, ct_2 = gi_tr_2 .

[WHERE iu_kin] Lu : cn thn vi cc lnh xa v sa, nu khng

c iu kin WHERE ngha l xa hoc sa tt c. V d: Tng gi 10% i vi nhng sn phm do

Trung Quoc sn xutUPDATE SANPHAMSET Gia = Gia*1.1WHERE Nuocsx=Trung Quoc

4.2 Sa d liu

4 3 Xa d liu


134/228


4.3 Xa d liu

C php

DELETE FROM tn_bng [WHERE iu_kin]

V d: Xa ton b nhn vin

DELETE FROM NHANVIEN

Xa nhng sn phm do Trung Quc sn xut c gi thp

hn 10000DELETE FROM SANPHAMWHERE (Gia


135/228


g g y cu trc

5.1 Cu truy vn tng qut5.2 Truy vn n gin

5.3 Php kt5.4 t b danh, s dng *, distinct5.5 Cc ton t5.6 Cu truy vn con (subquery)5.7 Php chia

5.8 Hm tnh ton, gom nhm

5 1Cu truy vn tng qut


136/228


SELECT [DISTINCT] *|tn_ct | hm

FROM bng

[WHERE iu_kin][GROUP BY tn_ct]

[HAVING iu_kin]

[ORDER BY tn_ct ASC | DESC]

5.1Cu truy vn tng qut

5 2 Truy vn n gin(1)


137/228


5.2 Truy vn n gin(1)

SELECT

Tng ng php chiu ca SQH Lit k cc thuc tnh cn hin th trong kt qu

WHERE

Tng ng vi iu kin chn trong SQH iu kin lin quan ti thuc tnh, s dng cc php ni

lun l AND, OR, NOT, cc php ton so snh,BETWEEN

FROM

Lit k cc quan h cn thit, cc php kt

5 2 Truy vn n gin(2)


138/228


5.2 Truy vn n gin(2)

Tm masp, tensp do Trung Quoc sn xutc gi t 20000 n 30000

Select masp,tensp

From SANPHAM

Where nuocsx=Trung Quocand gia between 20000 and 30000

5 3 Php kt(1)


139/228


5.3 Php kt(1)

Inner Join, Left Join, Right Join, Full Join

V d: In ra danh sch cc khch hng (MAKH,

HOTEN) mua hng trong ngy 1/1/2007.select KHACHHANG.makh,hotenfrom KHACHHANG inner join HOADON on

KHACHHANG.makh=HOADON.makhwhere nghd='1/1/2007'


140/228

5 4 t b danh s dng * distinct


141/228


5.4 t b danh, s dng , distinct

t b danh Alias: cho thuc tnh v quan h:tn_c AS tn_mi Select manv,hoten as [ho va ten] From NHANVIEN

Lit k tt c cc thuc tnh ca quan h: Select * from Nhanvien

Select NHANVIEN.* from NHANVIEN

Distinct: trng ch ly mt ln Select distinct nuocsx from SANPHAM

Sp xp kt qu hin th: Order by Select * from SANPHAM order by nuocsx, gia DESC

5 5 Ton t truy vn(1)


142/228


5.5 Ton t truy vn(1)

Ton t so snh: =, >,=,


143/228


5.5 Ton t truy vn(2)

IS NULL, IS NOT NULL

Select sohd from HOADON where makh is Null

Select * from HOADON where makh is Not Null

Ton t so snh, php ton Select gia*1.1 as [gia ban] from SANPHAM where

nuocsxViet Nam Select * from SANPHAM where (gia between 20000 and

30000) OR (nuocsx=Viet Nam) Ton t IN, NOT IN

Select * from SANPHAM where masp NOT IN(BB01,BB02,BB03)

5 5 Ton t so snh(3)


144/228


Ton t LIKE

So snh chui tng i

C php: s LIKE p, p c th cha % hoc _

% : thay th mt chui k t bt k

_ : thay th mt k t bt k V d: Select masp,tensp from SANPHAM

where masp like 'B%01

5.5 Ton t so snh(3)

5 6 Cu truy vn con (1)


145/228


5.6 Cu truy vn con (1)

In hoc Exists V d: Tm cc s ha n mua cng lc 2 sn phm

c m s BB01 v BB02. select distinct sohd

from CTHD where masp='BB01' and sohd IN(select distinct sohd from CTHD where masp='BB02')

select distinct A.sohd

from CTHD A where A.masp='BB01' andEXISTS (select * from CTHD B

where B.masp='BB02 and A.sohd=B.sohd)

5 6 Cu truy vn con (2)


146/228


5.6 Cu truy vn con (2)

Not In hoc Not Exists V d: Tm cc s ha n c mua sn phm m s

BB01 nhng khng mua sn phm m s BB02. select distinct sohd

from CTHD where masp='BB01' and sohd NOT IN(select distinct sohd from CTHD where masp='BB02')

select distinct A.sohd

from CTHD A where A.masp='BB01' andNOT EXITST (select * from CTHD B

where B.masp='BB02 and A.sohd=B.sohd)

5 7 Php chia


147/228


5.7 Php chia

S dng NOT EXISTS V d: Tm s ha n mua tt c nhng sn

phm do Trung Quoc sn xut.

Select sohd from HOADON where not exists(select * from SANPHAM

where nuocsx=Trung Quoc and not exists(select * from CTHD whereHOADON.sohd=CTHD.sohd andCTHD.masp=SANPHAM.masp))

5.8 Cc hm tnh ton v gomh (1)


148/228


5.8.1 Cc hm tnh ton c bn

COUNT: m s b d liu ca thuc tnh

MIN: Tnh gi tr nh nht

MAX: Tnh gi tr ln nht

AVG: Tnh gi tr trung bnh

SUM: Tnh tng gi tr cc b d liu

nhm (1)


149/228

V d


150/228


V d

1. Tnh lng thp nht, cao nht, trung bnh v tnglng ca tt c cc nhn vin.

2. C tt c bao nhiu nhn vin3. Bao nhiu nhn vin c ngi qun l

4. Bao nhiu phng ban c nhn vin trc thuc

5. Tnh lng trung bnh ca cc nhn vin

6. Tnh lng trung bnh ca cc nhn vin theotng phng ban


151/228


1. Tnh lng thp nht, cao nht, trung bnhv tng lng ca tt c cc nhn vin.

SELECT min(luong) as thapnhat,max(luong) as caonhat,avg(luong) as trungbinh,

sum(luong) as tongluong

FROM NhanVien


152/228


2. C tt c bao nhiu nhn vin

SELECT count(*) FROM NhanVien

3. Bao nhiu nhn vin c ngi qun l

Select count(*) FROM NhanVien WHERE manql is not null

SELECT count(Manql) FROM NhanVien

4. Bao nhiu phng ban c nhn vin trc thuc

SELECT count(distinct phong) FROM NhanVien



153/228


nhm (2)

5.8.2 Gom nhm: mnh GROUP BY

S dng hm gom nhm trn cc b trong quan h.

Mi nhm b bao gm tp hp cc b c cng gi tr trncc thuc tnh gom nhm

Hm gom nhm p dng trn mi b c lp nhau.

SQL c mnh GROUP BY ch ra cc thuc tnhgom nhm, cc thuc tnh ny phi xut hin trong mnh SELECT


154/228


5. Tnh lng trung bnh ca cc nhn vin

SELECT avg(LUONG) as LUONGTBFROM NhanVien

6. Tnh lng trung bnh ca cc nhn vin theo tngphng ban.

SELECT phong, avg(LUONG) as LUONGTBFROM NhanVienGROUP BY phong



155/228


nhm (3)

5.8.3 iu kin sau gom nhm: mnh HAVING Lc kt qu theo iu kin, sau khi gom nhm

iu kin HAVING c thc hin sau khi gom nhm,cc iu kin c lin quan n thuc tnh Group By

V d: tm phng c s lng nhn vin N trn 5 ngiSELECT phong

FROM NhanVienWHERE phai = N

GROUP BY phong

HAVING count(manv) > 5


156/228


Bi 6: Ngn ng tn t

Ni dung


157/228


N du g

1. Gii thiu2. C php

3. Cc nh ngha4. Din gii ca mt cng thc5. Quy tc lng gi cng thc

6. Ngn ng tn t c bin l n b7. Ngn ng tn t c bin l min gi tr

1. Gii thiu


158/228


Ngn ng tn t l ngn ng truy vn hnh thc do Codd ngh (1972-1973) c Lacroit, Proix v Ullman phttrin, ci t trong mt s ngn ng nh QBE, ALPHA..

c im: Ngn ng phi th tc Rt trch ci g ch khng phi rt trch nh th no Kh nng din t tng ng vi i s quan h

C hai loi: C bin l n b C bin l min gi tr

2. C php


159/228


p p

( ) : biu thc trong ngoc Bin: dng ch thng cui b k t: x,y,z,t,s Hng: dng ch thng u b k t: a,b,c, Hm: l mt nh x t mt min gi tr vo tp hp gm 2

gi tr: ng hoc sai. Thng dng ch thng gia bk t: h,g,f,

Tn t: l mt biu thc c xy dng da trn biu thc

logic. Dng ch in hoa gia b k t P,Q,R Cc php ton logic: ph nh (), ko theo (), v (),

hoc (). Cc lng t: vi mi (), tn ti ()

3. Cc nh ngha (1)


160/228


g ( )

nh ngha 1: Tn t 1 ngi Tn t 1 ngi c nh ngha trn tp X v bin x c gi tr

chy trn cc phn t ca X. Vi mi gi tr ca x, tn t P(x) l mt mnh logic, tc l nc gi tr ng () hoc sai (S) V d

P(x), x l bin chy trn X, l mt tn t

P(gt), gtX l mt mnh , X = {Nguyen Van A, Tran Thi B} Vi tn t N(x) c xc nh: x l ngi n. Khi Mnh N(Nguyen Van A): cho kt qu Sai N(Tran Thi B): cho kt qu ng

3. Cc nh ngha (2)


161/228


g ( )

nh ngha 2: Tn t n ngi Tn t n ngi c nh ngha trn cc tp X1, X2, , Xn v

n bin x1

, x2

, , xn

ly gi tr trn cc tp Xi

tng ng.

Vi mi gi tr aiXi, xi=ai.Tn t n ngi l mt mnh .

K hiu: P(x1, x2, , xn)

V d: CHA(x1,x2): x1l CHA ca x2

Ch : Cc Xikhng nht thit phi l ri nhau

Vi xi=ai, P(x1, x2, , ai, , xn) l tn t n-1 ngi

3. Cc nh ngha (3)


162/228


g ( )

nh ngha 3: T T l mt hng hay l mt bin Nu f(t1, t2, , tn) l hm n ngi th f l mt t

nh ngha 4: Cng thc Cng thc nguyn t: P(t1, t2, , tn), til cc t Nu F1, F2l cc cng thc th cc biu thc sau cng l cc

cng thc: F1F2, F1F2, F1=>F2, F1 Nu F1l mt cng thc th :F1, x:F1cng l cc cng thc Nu F1l cng thc th (F1) cng l mt cng thc

3. Cc nh ngha (4)


163/228


g ( )

nh ngha 4: Cng thc ngl cng thc nu mi bin u c km

vi lng t. (khng nh , S) Cng thc ml cng thc tn ti 1bin khng km

lng t. (tm kim thng tin)

V d: C1:xty(P(x,y,a)z(Q(y,z,t)R(x,t)) l cng thc

ng v cc bin x,y,z,t u c km lng t , C2:x t (P(x,y,a)z(Q(y,z,t)R(x,t)) l cng thc m

v bin y khng c lng t ,

4. Din gii ca mt cng thc


164/228


g g

Gm 4 phn: Min gi tr ca cc bin ca cng thc (k

hiu l tp M) S dng cc hng, cc tn t ( ngha tn t,

xc nh c quan h n ngi)

ngha ca cng thc Xc nh 1 quan h n ngi trn tp Mn

5. Quy tc lng gi cng thc


165/228


Q y g g g

Lng gi tn t: xt tn t bc n: P(x1,x2,xn) v linkt vi quan h R, n ngi.

P(a1,a2,,an): (a1,a2,,an) RP(a1,a2,,an): S (a1,a2,,an) R

Cc php ton ,,,dng bng chn tr Lng t : gi x l bin. Cng thc x F(x) l ng

khi c t nht mt ai

M/F(ai):M={a1,a2,,an} F(ai), aiM Lng t : x l bin, x F(x): vi aiM/F(ai):

M={a1,a2,,an} F(ai), aiM

6. Ngn ng tn t c bin l n b


166/228


g g

6.1 Qui tc

6.2 nh ngha

6.3 Cng thc an ton

6.4 Biu din cc php ton

6.1 Quy tc (1)


167/228


Q y ( )

1. Bin l 1b ca quan h2. T: hng, bin hoc biu thc c dng s[C], s:

bin, C: tp cc thuc tnh ca quan h c gil t chiu.

3. Cng thc: Rs (R l quan h, s l bin) c gi l t. Min gi tr

s nh ngha min bin thin ca s. t1 a , t1 t2 y t1,t2l cc t chiu, cn a l mt

hng, l ton t so snh dc gi l cng thcnguyn t

6.1 Quy tc (2)


168/228


Q y ( )

4. Mt cng thc nguyn tl mt cng thc

5. F1 v F2l cng thc: F1F2, F1F2,

F1F2, F1 l cng thc6. F l cng thc , s l bin sF, sF l cng

thc

7. F l cng thc, (F) l cng thc

6.2 nh ngha


169/228


g

Mt cu hi trong ngn ng tn t c bin ln b c biu din nh sau: {s | F} . Trong

s l bin n b, F l mt cng thc ch cmt bin t do l s. V d: BIENGIOI(nuoc,tinhtp). Php ton

quan h BIENGIOI[nuoc] c chuyn

thnh cu hi trong ngn ng tn t c binl b: {s[nuoc] BIENGIOI s}

6.3 Cng thc an ton


170/228


F l cng thc an ton: nu n tho mn 3 iu kin sau:i) Nu s l b n tha: F(s) l ng th mi thnh phn ca s

l phn t ca DOM(F):

ii) F l cng thc con ca F:

iii)

g

)():( FDOMsngsF

)'(:',' FDOMsngsFssF

)'(:',' FDOMsngsFssF

6.4 Biu din cc php ton (1)


171/228


1. Php hi Q1,Q2l cc quan h n chiu F1, F2l cc cng thc ng vi Q1, Q2 Cng thc ca Q= Q1Q2 Fs=F1sF2s

2. Php tr Q

1

,Q2

l cc quan h n chiu F1, F2l cc cng thc ng vi Q1, Q2 Cng thc ca Q= Q1-Q2 Fs=F1F2s



172/228


3. Php tch

Q1(x1,,xm), Q2(y1,,yn)

F1, F2l cc cng thc ng vi Q1, Q2 Cng thc ca Q= Q1 x Q2

Fs: s(x1,,xm,y1,,yn)

Fs=(v) ( p) (F1v F2p s1=v1sm=vm sm+1=p1 sm+n=pn)



173/228


4. Php chiu Q1(x1,,xn), F1l cc cng thc ng vi Q1

Cng thc ca Q= Q1 [xi1, xi2,,xik]Fs=(v) (F1v s1=vi1s2=vi2 sk=vik)

5. Php chn Q1l quan h n chiu, F1l cng thc ng vi Q1 Cng thc Q=Q1:iu kin K (K:xixjhoc xia)

Fs=F1s si sj hoc F1s si a (1i, j n, ij)

7. Ngn ng tn t c bin l min gitr


174/228


tr

7.1 Quy tc

7.2 Biu din cu hi

7.3 Cng thc an ton

7.4 Biu din cc php ton

7.1 Quy tc


175/228


1. T: l hng hoc bin2. Cng thc nguyn t

Q(t1

,t2

,,tn

): ti

l cc t, Q l quan h ti tj ,tia vi til t, a l mt hng, l php ton

3. Mt cng thc nguyn tl mt cng thc4. F1 v F2l cng thc: F1F2, F1F2, F1F2, F1 l

cng thc5. F l cng thc , t:bin t do, sF,sF cng cng thc6. F l cng thc, (F) l cng thc

7.2 Biu din cu hi


176/228


{(x1,x2,,xn) | F(x1,x2,,xn)}

xil cc bin t do ca F

Q= {(x1,x2,,xn) | F(x1,x2,,xn)} nn(x1,x2,,xn)Q F(x1,x2,,xn):ng

7.3 Cng thc an ton


177/228


F l cng thc an ton: nu n tho mn 3 iu kin sau:i) Nu s l b n tha: F(s) l ng th mi thnh phn ca s

l phn t ca DOM(F):

ii) F l cng thc con ca F:

iii)

niFDOMi

xngnxxF ,...,1,)():),...,1((

)'(:' FDOMxngxF

)'(:' FDOMxngxF

niFDOMi

xngnxxF ,...,1,)():),...,1((



178/228


1. Php hi Q1,Q2l cc quan h n chiu F1, F2l cc cng thc ng vi Q1, Q2 Cng thc ca Q= Q1Q2 F=F1F2

2. Php tr Q

1

,Q2

l cc quan h n chiu F1, F2l cc cng thc ng vi Q1, Q2 Cng thc ca Q= Q1-Q2 F=F1F2



179/228


3. Php tch

Q1(x1,,xm), Q2(y1,,yn)

F1, F2l cc cng thc ng vi Q1, Q2 Cng thc ca Q= Q1 x Q2

F(x1,,xm,y1,,yn) =F1(x1,,xm)F2(y1,,yn)



180/228


4. Php chiu Q1(x1,,xn), F1(x1,,xn) l cc cng thc ng vi Q1 Cng thc ca Q= Q1 [xi1, xi2,,xik]

Fs(xi1, xi2,,xik)= (xji)(xjz)(xjn-k)(F1(x1,,xn))trong (xi1, xi2,,xik)(xj1, xj2,,xjn-k)=(x1, x2,,xn)

5. Php chn Q1(x1,,xn), F1(x1,,xn) l cc cng thc ng vi Q1 Cng thc Q=Q1:iu kin K (K:xixjhoc xia)

F1(x1,,xn)= F1(x1,,xn) xi xj hoc= F1(x1,,xn) xi a

Bi 7: Rng buc ton vn


181/228


Bi 7: Rng buc ton vn

Ni dung chnh


182/228


1. Gii thiu rng buc ton vn (RBTV)

2. Cc c trng ca mt RBTV

3. Phn loi RBTV4. Bng tm nh hng tng hp

1. Gii thiu


183/228


Rng buc ton vn l cc quy nh, iu kin tng dng thc t, cc iu kin ny l bt bin.

V th phi lun m bo c s d liu tho rngbuc ton vn sau mi thao tc lm thay i tnhtrng ca c s d liu.

2. Cc c trng ca mtRBTV


184/228


2.1 Ni dung

2.2 Bi cnh

2.3 Bng tm nh hng

RBTV

2.1 Ni dung


185/228


M t cht ch ngha ca rng buc tonvn.

Ni dung c pht biu bng ngn ng tnhin hoc bng ngn ng hnh thc (ngnng tn t, i s quan h, m gi,) Ngn ng t nhin: d hiu nhng khng cht

ch, logic. Ngn ng hnh thc: cht ch, c ng

2.2 Bi cnh


186/228


L tp cc quan h khi thao tc trn nhngquan h c kh nng lm cho rng buc b

vi phm. l nhng quan h c th vi phm rng

buc ton vn khi thc hin cc thao tc

thm, xo, sa.

2.3 Bng tm nh hng (1)


187/228


Nhm xc nh khi no tin hnh kim trarng buc ton vn. Thao tc no thc hin

c th lm vi phm rng buc ton vn. Phm vi nh hng ca mt rng buc ton

vn c biu din bng mt bng 2 chiu

gi l bng tm nh hng.



188/228


Mt s quy nh Nhng thuc tnh kho (nhng thuc tnh nm

trong kho chnh ca quan h) khng c phpsa gi tr

Thao tc thm v xo xt trn mt b ca quan h.Thao tc sa xt sa tng thuc tnh trn b caquan h

Trc khi xt thao tc thc hin c th lm vi phmrng buc hay khng th CSDL phi tho rng bucton vn trc.



189/228


Rng bucRi

Thm Xa Sa

Quan h 1

Quan h n

Bng tm nh hng ca mt rng buc

+ : thc hin thao tc c th lm vi phm RBTV- : thc hin thao tc khng th lm vi phm RBTV+(A) : c th lm vi phm RBTV khi sa trn thuc tnh A

(*) : khng vi phm RBTV do thao tc khng thc hin c

3. Phn loi


190/228


3.1 RBTV c bi cnh trn 1 quan h

3.2 RBTV c bi cnh trn nhiu quan h

3.3 Ph thuc hm (functional dependency)

3.1 RBTV c bi cnh 1 quan h


191/228


3.1.1 RBTV min gi tr.

3.1.2 RBTV lin thuc tnh

3.1.3 RBTV lin b

Lc CSDL qun l gio v


192/228



LOP(MALOP, TENLOP, TRGLOP, SISO, MAGVCN)

KHOA(MAKHOA, TENKHOA, NGTLAP, TRGKHOA)

MONHOC(MAMH, TENMH, TCLT, TCTH, MAKHOA)






3.1.1 Rng buc min gi tr


193/228


L tp gi tr m mt thuc tnh c th nhn.

R1:Gii tnh ca hc vin ch l Nam hoc N

Ni dung:hv HOCVIEN: hv.Gioitinh {Nam,N}

Bi cnh: quan h HOCVIEN

Bng tm nh hng:

R1 Thm Xa Sa

HOCVIEN + - +(Gioitinh)

3.1.2 Rng buc lin thuc tnh


194/228


L rng buc gia cc thuc tnh vi nhau trn 1 bca quan h

R2:Ngy bt u (TUNGAY) ging dy mt mn hc cho mt

lp lun nh hn ngy kt thc (DENNGAY)

Ni dung:gd GIANGDAY: gd.TUNGAY < gd.DENNGAY

Bicnh : GIANGDAY

Bng tm nh hng:

R2 Thm Xa Sa

GIANGDAY + - +(Tungay, Denngay)

3.1.3 Rng buc lin b (1)


195/228


L rng buc gia cc b trn cng mt quan h (c thlin quan n nhiu thuc tnh).

R3: Tt c cc hc vin phi c m s phn bit vi nhau

Ni dung:h1,h2HOCVIEN: Nu h1h2 th h1.Mahvh2.Mahv Bi cnh: quan h HOCVIEN Bng tm nh hng:

R3 Thm Xa Sa

HOCVIEN + - -(*)


196/228

3.2 RBTV c bi cnh nhiu quanh


197/228


3.2.1 RBTV tham chiu (kho ngoi, phthuc tn ti)

3.2.2 RBTV lin thuc tnh3.2.3 RBTV do thuc tnh tng hp

3.2.4 RBTV do chu trnh trong lc biu

din quan h

3.2.1 Rng buc tham chiu (1)


198/228


L rng buc quy nh gi tr thuc tnhtrong mt b ca quan h R (tp thuc tnh

ny gi l kho ngoi), phi ph thuc vo stn ti ca mt b trong quan h S (tp thuctnh ny l kho chnh trong quan h S).

RBTV tham chiu cn gi l rng buc phthuc tn ti hay rng buc kha ngoi

3.2.1 Rng buc tham chiu (2)


199/228


R5: Hc vin thi mt mn hc no th mn hc phi c trong danh sch cc mn hc

Ni dung:

k KETQUATHI, m MONHOC: k.Mamh = m.Mamh Hoc: KETQUATHI[Mamh] MONHOC[Mamh]

Bi cnh: quan h KETQUATHI, MONHOC

Bng tm nh hng:

R5 Thm Xa Sa

KETQUATHI + - -(*)

MONHOC - + -(*)

3.2.2 Rng buc lin thuc tnh (1)


200/228


L rng buc gia cc thuc tnh trn nhng quan hkhc nhau

R6: Ngy gio vin ging dy mt mn hc phi ln hn hoc

bng ngy gio vin vo lm. Ni dung: gd GIANGDAY

Nu gv GIAOVIEN: gd.Magv = gv.Magv thgv.NGVL gd.TUNGAY

Bi cnh: GIANGDAY, GIAOVIEN Bng tm nh hng: R6 Thm Xa Sa

GIANGDAY + - +(Tungay)

GIAOVIEN - - +(Ngvl)



201/228


R7:Ngy thi mt mn hc phi ln hn ngy kt thc hcmn hc .

Ni dung:

kq KETQUATHINu gd GIANGDAY, hv HOCVIEN:

(gd.Malop=hv.Malop)(kq.Mamh=gd.Mamh) th

gd.Denngay < kq.Ngthi

Bi cnh: GIANGDAY, HOCVIEN, KETQUATHI



202/228


Bng tm nh hng:

R7 Thm Xa Sa

HOCVIEN - - +(Malop)

GIANGDAY - - +(Denngay)

KETQUATHI + - +(Ngthi)


203/228

3.2.3 RBTV do thuc tnh tng hp (2)


204/228


Doanh s ca mt khch hng bng tng tr gi ccho n m khch hng mua Ni dung:

kh KHACHHANG,

kh.Doanhso = (hd HOADON: hd.Makh=kh.Makh)(hd.Trigia)

Bi cnh: KHACHHANG, HOADON

Bng tm nh hng:

Thm Xa Sa

KHACHHANG + - +(Doanhso)

HOADON + + +(Makh)

3.2.3 RBTV do thuc tnh tng hp (3)


205/228


R8: S s ca mt lp l s lng hc vin thuc lp

Ni dung:

l LOP,

l.Siso = Count(hv HOCVIEN: hv.Malop = lp.Malop)(*)

Bi cnh: quan h LOP, HOCVIEN

Bng tm nh hng:

R8 Thm Xa Sa

LOP + - +(Siso)

HOCVIEN + + +(Malop)

3.2.4 Do hin din ca chu trnh (1)


206/228


Biu din lc quan h di dng th: Quan h c biu din bng nt trn rng to

Thuc tnh c biu din bng nt trn c nh Tt c cc nt u c ch r bng tn ca quan h

hoc thuc tnh. Thuc tnh thuc mt quan hc biu din bi mt cung ni gia nt trn to v

nt trn nh Nu th biu din xut hin mt ng khp kn

=> lc CSDL c s hin din ca chu trnh.



207/228


GIAOVIEN

GIANGDAY

MONHOCTenmh

TCLT

Mamh

MalopMagv

Hoten

Hocvi

Makhoa

Y

X



208/228


X = GIANGDAY[Magv, Mamh]

Y = (GIAOVIENMONHOC) [Magv,Mamh]

ngha: X: gio vin v nhng mn hc c phn cng cho

gio vin ging dy

Y: gio vin v nhng mn hc thuc khoa gio vin ph trch

Mi quan h gia X v Y trong cc rng buc sau:

Makhoa



209/228


Rng buc 1: gio vin ch c phn cngging dy nhng mn thuc khoa gio vin ph trch XY

Rng buc 2: gio vin phi c phn cngging dy tt c nhng mn thuc khoa giovin ph trch X=Y

Rng buc 3: c th phn cng gio vinging dy bt k mn hc no X Y



210/228


R9: gio vin ch c phn cng ging dy nhngmn thuc khoa gio vin ph trch XY

R9 Thm Xa Sa

MONHOC - - +(Makhoa)

GIAOVIEN - - +(Makhoa)

GIANGDAY + - +(Magv)

3.3 Ph thuc hm (1)


211/228


Cho quan h Q(A, B, C). Ph thuc hm A xc nh

B. K hiu A B nu:

q1,q2Q: Nu q1.A=q2.A th q1.B=q2.B

A B c gi lph thuc hm hin nhinnu

BA

A B c gi lph thuc hm nguyn tnuAA, AA sao cho A B

3.3 Ph thuc hm (2)


212/228


Mi quan h u c t nht mt ph thuc hm

Rng buc kho cng l mt ph thuc hm

Mamh Tenmh, Tclt, Tcth, Makhoa R4: Cc gio vin c cng hc v, cng h s lng

th mc lng s bng nhau. Rng buc ny c th

biu din bng ph thuc hm nh sau:Hocvi,Heso Mucluong

4. Bng tm nh hng tng hp (1)


213/228


Bng tm nh hng tng hpca m rng buc trn nquan h bi cnh

QH1

QH2

QHn

T X S T X S T X S

R1

R2

Rm

4. Bng tm nh hng tng hp (2)


214/228


HOCVIEN GIAOVIEN LOP MONHOC GIANGDAY KETQUA

THI

T X S T X S T X S T X S T X S T X S

R1 + - +

R2 + - +

R3 + - -*

R4 + - +

R5 - + -* + - -*

R6 - - + + - +R7 - - + - - + + - +

R8 + + + + - +

R9 - - + - - + + - +

Bi 8: Ti u ha cu hi


215/228


Ni dung


216/228


1. Gii thiu2. Cc nguyn tc tng qut ti u ha cu hi

2.1 Biu thc tng ng

2.1.1 nh ngha2.1.2 Tnh cht ca php kt v php tch

2.2 Nguyn tc tng qut2.3 Cc php bin i tng ng

3. Mt s k thut ti u ha cu hi bng SQH3.1 K thut (dy php chn, php chiu, hon v )3.2 Thut gii ti u ho cu hi trong .

1. Gii thiu (1)


217/228


Mc ch: Gim thi gian x l cu hi, gim khi lng

d liu trung gian. Kt hp gia cc php tch, php kt vi php

chn vi php chiu.

V d:

])[):((

])[:)((

201

021

CQaAQ

CaAQQ

1. Gii thiu (2)


218/228


K hiu:

X

R

Q

D

R

Q

AB

R S

Q

Q=R[S]

Q=R:D

Q=R S

BA

1. Gii thiu (3)


219/228


V d

Q1 Q2

AA=a0

C

A

Q1

C

Q2A=a0

])[:)(( 021 CaAQQ ])[):(( 201 CQaAQ

2.1 Tnh tng ng (1)


220/228


2.1.1 nh ngha: hai biu thc A, B l tngng nu c cng mt tnh trng CSDL th ucho mt kt qu.

2.1.2 Tnh cht ca php kt v php tch Php kt

Giao hon

Kt hp

Php tch

Giao hon:

Kt hp:

1221 QQQQdkdk

3

2

2

1

13

2

2

1

1 )()( QQQQQQdkdkdkdk

1221 QQQQ

321321 )()( QQQQQQ

2.1 Tnh tng ng (2)


221/228


2.1.3 Cc php bin i tng ng

]))[,(][][((][),(),(.5

),...,(])[...][][(),...,(.4

))()((.3

):(),(),(.2

])[][:(),(),(.1

121121

1211

2121

2121

212121

BBAQAQBQBQBAQBAQ

XXQXQXQXQXXQ

QQQQ

DBQQDCQBAQ

BQBQQQCBQBAQ

nnn

DB

B

2.2 Nguyn tc tng qut


222/228


1. Thc hin php chiu, php chn cng sm cng tt

2. Gom cc php chn v chiu cng quan h thc

hin cng lc3. Bin php tch thnh php kt t nhin hay theta kt

4. Tm cc biu thc con chung trong mt biu thc

5. Tin x l cc quan h: lp ch mc6. nh gi trc khi thc hin tnh ton

3.1 Cc k thut ti u (1)


223/228


1. Dy cc php chn2. Dy cc php chiu3. Hon v gia php chiu v php chn4. Hon v gia php chn v php tch5. Hon v gia php hp v php chn6. Hon v gia php chn v php tr

7. Hon v gia php chiu v php hi8. Hon v gia php chiu v php tch



224/228


1. Dy cc php chn

2. Dy php chiu

V d:

dkndkdkQdkndkdkQ ...21:):)...2:)1:(((

YZZQZYQ ,][]])[[(

][]])[,,[(

),,,(

ADQADDCAQ

DCBAQCho



225/228


3. Hon v gia php chiu v php chn Nu

Nu YX

YX

)(:])[(]))[(:( XdkYXQYXdkQ

)(:])[(]))[(:( XdkYQYXdkQ



226/228


4. Hon v gia php chn v php tch: iu kin dk xc lp trn cc thuc tnh ca X

Nu , dk1 xc lp trn cc thuctnh ca X, dk2 xc lp trn cc thuc tnh ca Y.

Nu dk1 xc lp trn cc thuc tnh ca X v dk2xc lp trn cc thuc tnh ca XY

dkYQXQYQXdkXQ :))()(()()(:))(( 2121

21 dkdkdk

)2:)(()1:)((()(2)(1:))()((( 2121 dkYQdkXQYdkXdkYQXQ

))(2:))(()1:)(((

)(2)(1:))()(((

21

21

YXdkYQdkXQ

YXdkXdkYQXQ



227/228


5. Hon v gia php hi v php chn

6. Hon v gia php chn v php tr

7. Hon v gia php chiu v php hi

8. Hon v gia php chiu v php tch

):():(:)( 2121 dkQdkQdkQQ

):():(:)( 2121 dkQdkQdkQQ

])[(])[(])[( 2121 ZQZQZQQ

YXZZYQZYQZYQXQ ,])[(])[(]))[()(( 2121

3.2 Thut ton


228/228

Bc 1: p dng cc php bin i tng ng Bc 2: p dng (1) Bc 3: i vi cc php chn p dng (3), (4), (5), (6)

nhm a php chn cng su cng tt Bc 4: i vi cc php chiu p dng (2), (3), (7), (8)

nhm a php chiu cng su cng tt

Documents

co so du lieu SQL