Database Redesign. 1. Additional SQL Statements 3 1) Correlated Sub-Query 3 2) EXISTS 4 3) NOT EXISTS 7 4) double NOT EXISTS (FOR ALL) 9.

Transcription

1 Database Redesign 1. Additional SQL Statements 3 1) Correlated Sub-Query 3 2) EXISTS 4 3) NOT EXISTS 7 4) double NOT EXISTS (FOR ALL) 9 [Report] 14

2 Additional SQL Statements (Database Redesign) What we need to learn in chapter 8 1. Database Redesign 2. Additional SQL Statement - Exists - Not Exists 3. Analyzing the Existing Database 4. Changing Table Names and Table Columns 5. Changing Relationship Cardinalities and Properties

3 1. Additional SQL Statements 1) Correlated Sub-Query The following correlated subquery can be used to check for any rows that violate the functional dependency Department BudgetCode SELECT FROM WHERE E1.Department, E1.BudgetCode EMPLOYEE E1 E1.Department IN (SELECT E2.Department FROM EMPLOYEE E2 WHERE E1.Department = E2.Department AND E1.BudgetCode <> E2.BudgetCode); E1 E2

4 2) EXISTS SELECT FROM WHERE E1.Department, E1.BudgetCode EMPLOYEE E1 EXISTS (SELECT * FROM EMPLOYEE E2 WHERE E1.Department = E2.Department AND E1.BudgetCode <> E2.BudgetCode); (EX) x = 1, 5, 10 y = 5, 10, 15 Find an integer x such that there exists an integer y such that x > y y ( x > y )

5 S P S# S2 S3 S5 P# P1 P3 P4 P5 P6 SNAME Smith Jones Blakes Clark Adams PNAME Nut Bolt Screw Screw Cam Cog STATUS COLOR Red Green Blue Red Blue Red CITY London Paris Paris London Athenes WEIGHT CITY London Paris Rome London Paris London SP S# S2 S2 S3 P# P1 P3 P4 P5 P6 P1 P4 P5 QTY 부품번호가 p2" 인부품을공급하는공급자이름을구하시오 SELECT DISTINCT S.SNAME FROM S WHERE EXISTS ( SELECT * FROM SP WHERE SP.S# = S.S# AND SP.P# = '' ) ; Find a S such that there exists a SP such that (SP.S# = S.S# and SP.P#='') Find a S such that sp, (SP.S# = S.S# and SP.P#='') is equal to SELECT DISTINCT S.SNAME FROM S WHERE S.S# IN ( SELECT SP.S# FROM SP WHERE SP.P# = '' ) ; is equal to SELECT DISTINCT S.SNAME FROM S, SP WHERE SP.S# = S.S# AND SP.P# = '' ) ;

6 -- By C program #include <stdio.h> #define SMAX 5 #define PMAX 6 #define SPMAX 12 int S[SMAX] = 1, 2, 3, 4, 5; int SP[SPMAX][2] = 1,1, 1,2, 1,3, 1,4, 1,5, 1,6, 2,1, 2,2, 3,2, 4,2, 4,4, 4,5 ; main() int i,j,k; for(i=0; i<smax; i++) for(k=0; k<spmax; k++) if ( ( S[i] == SP[k][0] ) && ( SP[k][1] == 2 ) ) printf(" S : %5d \n", S[i]); continue;

7 3) NOT EXISTS 부품번호 p2" 를공급하지않는공급자이름을구하시오 SELECT DISTINCT S.SNAME FROM S WHERE NOT EXISTS ( SELECT * FROM SP WHERE SP.S# = S.S# AND SP.P# = '' ) ; is equal to SELECT DISTINCT S.SNAME FROM S WHERE S.S# NOT IN (SELECT SP.S# FROM SP WHERE SP.P# = '' ) ; S S# S2 S3 S5 SNAME Smith Jones Blakes Clark Adams STATUS CITY London Paris Paris London Athenes SP S# S2 S2 S3 P# P1 P3 P4 P5 P6 P1 P4 P5 QTY

8 -- #include <stdio.h> #define SMAX 5 #define PMAX 6 #define SPMAX 12 int S[SMAX] = 1, 2, 3, 4, 5; int SP[SPMAX][2] = 1,1, 1,2, 1,3, 1,4, 1,5, 1,6, 2,1, 2,2, 3,2, 4,2, 4,4, 4,5 ; int N_EXISTS[SMAX] = 0 ; main() int i,j,k; int NOTEXISTS = 1; for(i=0; i<smax; i++) for(k=0; k<spmax; k++) if ( ( S[i] == SP[k][0] ) && ( SP[k][1] == 2 ) ) printf(" S : %5d \n", S[i]); NOTEXISTS = 0 ; continue; if ( NOTEXISTS ) N_EXISTS[i] = 1 ; NOTEXISTS = 1; for ( i=0; i<smax; i++) if ( N_EXISTS[i] ) printf(" Supplier : %d\n", S[i]);

9 4) double NOT EXISTS (FOR ALL) 모든부품을공급하는공급자이름을구하시오 SELECT DISTINCT S.SNAME FROM S WHERE NOT EXISTS (SELECT * FROM P WHERE NOT EXISTS (SELECT * FROM SP WHERE SP.S# = S.S# AND SP.P# = P.P# ) ) ; FORALL in terms of negated existential quantifiers FORALL x (f) = NOT EXISTS x ( NOT f) "all x's satisfy f" is the same as "no x's do not satisfy f" FORALL x ( EXIST y (y>x) = NOT EXISTS x ( NOT EXISTS y (y>x) "For all integer x, there exists an integer y such that y > x" is equivalent to the statement (EX) x = 1, 5, 10 y = 5, 10, 15 "there does not exist an integer x such that there does not exist an integer y such that y > x" FORALL P (EXIST SP (... ) = NOT EXIST P (NOT EXIST SP (... ) )

10 가. part 1 ( 공급하고있는상황에대한모든정보 ) select * from SP, s, p where s.s# = sp.s# and p.p# = sp.p# (12 개행적용됨 ) 나. part 1과같은표현 select * from s, p where EXISTS ( select * from SP where s.s# = sp.s# and p.p# = sp.p# ) ;

11 다. part 2 /* all combinations of s, sp, p that are not satisfied join condition */ ( 공급안하고있는모든상황 ) select * from s, p where NOT EXISTS ( select * from SP where s.s# = sp.s# and p.p# = sp.p# ) ; 여기에만족하는않는공급자? = select * from s where NOT EXISTS ( select * from p where NOT EXISTS ( select * from SP where s.s# = sp.s# and p.p# = sp.p# ) ) ;

12 [Program to illustrate the SQL NOT EXISTS] #include <stdio.h> #define SMAX 5 #define PMAX 6 #define SPMAX 12 int S[SMAX] = 1, 2, 3, 4, 5; int P[PMAX] = 1, 2, 3, 4, 5, 6; int SP[SPMAX][2] = 1,1, 1,2, 1,3, 1,4, 1,5, 1,6, 2,1, 2,2, 3,2, 4,2, 4,4, 4,5 ; int N_EXISTS[SMAX] = 0 ; main() int int i,j,k; NOTEXISTS=1; for(i=0; i<smax; i++) for ( j=0; j<pmax; j++) for(k=0; k<spmax; k++) if ( ( S[i] == SP[k][0] ) && ( P[j] == SP[k][1] ) ) NOTEXISTS = 0; printf("matched SP S# : %d P# : %d \n", SP[k][0], SP[k][1] ); continue; if ( NOTEXISTS ) printf(" S : %5d, P : %5d \n", S[i], P[j] ); N_EXISTS[i] = 1; /* constructing the result of PART 2 */ NOTEXISTS = 1; for ( i=0; i<smax; i++) if (! N_EXISTS[i] ) printf(" Supplier Number : %d\n", S[i] );

13 The following code determines the name of any ARTIST that is of interest to every CUSTOMER: SELECT FROM WHERE A.Name ARTIST AS A NOT EXISTS (SELECT C.CustomerID FROM CUSTOMER C WHERE NOT EXISTS (SELECT CI.CustomerID FROM CUSTOMER_artist_int CI WHERE C.CustomerID = CI.CustomerID AND A.ArtistID = CI.ArtistID));

14 [Report ] Implement the C program code such like one in page 12 corresponding to the SQL code in page 13