记一次数据库课后作业：SQL查询应用
作者：互联网
本文仅学习使用
创建关系
仅简单创建了几个关于工程的关系
create table S --创建供应商关系
( S# varchar(10), --供应商号码
SNAME varchar(10), --供应商名字
SCITY varchar(10), --供应商城市
primary key(S#)); --主码为S#
insert into S
values('S1','东','Nanjing');
insert into S
values('S2','南','Shanghai');
insert into S
values('S3','西','Beijing');
insert into S
values('S4','北','Shanghai');
create table P --创建零件关系
( P# varchar(10), --零件号
PNAME varchar(10), --零件名字
COLOR varchar(10), --零件颜色
WEIGHT smallint, --零件重量
primary key(P#));
insert into P
values('P1','壹','red',10);
insert into P
values('P2','贰','yellow',18);
insert into P
values('P3','叁','blue',20);
insert into P
values('P4','肆','red',22);
create table J --创建工程关系
( J# varchar(10), --工程号
JNAME varchar(10), --工程名
JCITY varchar(10), --工程所在城市
primary key(J#));
insert into J
values('J1','一','北京');
insert into J
values('J2','二','上海');
insert into J
values('J3','三','天津');
insert into J
values('J4','四','南京');
create table SPJ --创建SPJ关系
( S# varchar(10),
P# varchar(10),
J# varchar(10),
QTY varchar(20), --供应商、零件、工程三者的关系
primary key(S#,P#,J#),
foreign key(S#) references S,
foreign key(P#) references P,
foreign key(J#) references J,);
INSERT INTO SPJ(S#,P#,J#,QTY) VALUES('S1','P1','J1',200);
INSERT INTO SPJ(S#,P#,J#,QTY) VALUES('S1','P1','J3',100);
INSERT INTO SPJ(S#,P#,J#,QTY) VALUES('S1','P1','J4',700);
INSERT INTO SPJ(S#,P#,J#,QTY) VALUES('S1','P2','J2',100);
INSERT INTO SPJ(S#,P#,J#,QTY) VALUES('S2','P3','J1',400);
INSERT INTO SPJ(S#,P#,J#,QTY) VALUES('S2','P3','J2',200);
INSERT INTO SPJ(S#,P#,J#,QTY) VALUES('S2','P3','J4',500);
INSERT INTO SPJ(S#,P#,J#,QTY) VALUES('S2','P3','J3',400);
INSERT INTO SPJ(S#,P#,J#,QTY) VALUES('S2','P4','J1',400);
INSERT INTO SPJ(S#,P#,J#,QTY) VALUES('S2','P4','J2',100);
a.找出给project J1 和J2都提供零件的所有供应商号码
select S#
from SPJ
where J#='J1' AND J#='J2'; --写法1
(select S#
FROM SPJ
WHERE J#='J1')
INTERSECT
(select S#
FROM SPJ
WHERE J#='J2') ; --写法2
select distinct S#
FROM SPJ
WHERE J#='J1' AND
S# IN(SELECT S#
FROM SPJ
WHERE J#='J2'); --写法3
b.找出为北京或上海的任何工程提供了零件的全部供应商号码
select S#
FROM SPJ ,J
where JCITY='Beijing'or JCITY='Shanghai'and SPJ.J#=J.J#; --写法1
(select S#
FROM SPJ ,J
where JCITY='Beijing' AND SPJ.J#=J.J#)
UNION
(select S#
FROM SPJ ,J
where JCITY='Shanghai' AND SPJ.J#=J.J#); --写法2
select S#
FROM SPJ,J
WHERE JCITY='Beijing' and SPJ.J#=J.J# or
S# IN (SELECT S#
FROM SPJ,J
WHERE JCITY='Beijing' and SPJ.J#=J.J#); --写法3
c 找出所有为上海的工程提供零件的所有上海的供应商号码
SELECT S.S#
from S,J,SPJ
WHERE J.J#=SPJ.J# and S.S#=SPJ.S# AND JCITY='Shanghai' and SCITY='Shanghai'; --写法1
(SELECT S.S#
FROM S,J,SPJ
WHERE J.J#=SPJ.J# and S.S#=SPJ.S# AND JCITY='Shanghai' )
INTERSECT
(SELECT S.S#
FROM S,J,SPJ
WHERE J.J#=SPJ.J# and S.S#=SPJ.S# AND SCITY='Shanghai'); --写法2
SELECT S.S#
FROM S,J,SPJ
WHERE J.J#=SPJ.J# and S.S#=SPJ.S# AND JCITY='Shanghai'AND
S.S# IN(SELECT S.S#
FROM S,J,SPJ
WHERE J.J#=SPJ.J# and S.S#=SPJ.S# AND SCITY='Shanghai'); --写法3
d.找出从不提供红色零件的北京供应商号码
(select S.S#
FROM S
WHERE SCITY='Beijing')
except
(select SPJ.S#
FROM SPJ,P
WHERE SPJ.P#=P.P# AND COLOR='red'); --写法1
select S.S#
FROM S
WHERE SCITY='Beijing' AND
S.S# NOT IN (select SPJ.S#
FROM SPJ,P
WHERE SPJ.P#=P.P# AND COLOR='red'); --写法2
select S.S#
FROM S
WHERE SCITY= 'Beijing' and
not exists(select SPJ.S#
FROM SPJ,P
WHERE SPJ.P#=P.P# AND COLOR='red'); --写法3
e 找出至少使用供应商S1提供的零件的工程的工程号码://只要用过s1提供过的零件就可以(题目理解1）
select distinct J#
FROM SPJ
WHERE S#='S1';
–e找出至少使用供应商S1所供应的全部零件的工程号码：（题目理解2）
–间接法：不存在这样的零件y：供应商s1供应了y,而工程没有选择y（不存在工程使用y);
select distinct J#
FROM SPJ AS SPJ3 --第三步：下一行的not exists表示若这种情况不存在则这个工程符合
where not exists (select * --第一步找出S1供应的所有零件
from SPJ AS SPJ1
WHERE S#='S1'and not exists(select* --第二步找出其他工程使用过的S1零件，取不存在：找出其他工程没有使用过S1提供的零件
from SPJ AS SPJ2
where SPJ1.P#=SPJ2.P#
AND SPJ2.J#=SPJ3.J#)); --返回该工程编号，让SPJ3接收
–直接法：首先找出S1供应商供应的全部零件集合P1
/*
select A.P#
FROM SPJ AS A
where A.S#='S1';
--再对某一个工程找出其使用的零件集合Pn
SELECT B.P#
FROM SPJ AS B
WHERE B.J#='某工程的工程号'
--如果P1⊆Pn，则说明该工程使用了S1供应商供应的全部零件
--只要求P1-Pn，若P1-Pn为空集（查询结果不存在），则说明P1⊆Pn。
select *
from (SELECT A.P#
FROM SPJ AS A --P1
WHERE A.S#='S1')AS C
WHERE C.P# NOT IN(SELECT B.P# -- -Pn
FROM SPJ AS B
WHERE B.J#='某工程的工程号' --返回为空则说明P1⊆Pn
*/
SELECT D.J# --用D.J#代替B.J#='某工程的工程号' 的后者的占位，遍历J表中的每一个工程
FROM J AS D
WHERE NOT EXISTS(SELECT*
FROM (SELECT A.P#
FROM SPJ AS A
WHERE A.S#='S1')AS C
WHERE C.P# NOT IN(SELECT B.P#
FROM SPJ AS B
WHERE B.J#=D.J#));
–f.找出提供所有零件的供应商姓名
–直接法：找出P关系中的所有零件集合P1,找出某一供应商供应的所有零件集合Pn
–若P1⊆Pn（P1-Pn查询结果不存在），则找出了提供所有零件的供应商
select SNAME
FROM S
WHERE NOT EXISTS(select A.P#
from P AS A
WHERE A.P# NOT IN(SELECT B.P#
FROM SPJ AS B
WHERE B.S#=S.S#));
–g 找出这样的供应商的号码：供应商S1能提供的零件该供应商都能提供
–令S1所提供的零件集合为P1，该供应商能提供的零件为Pn
–若P1⊆Pn（P1-Pn查询结果不存在），则找出了S1能提供的零件该供应商都能提供的供应商
select S#
FROM S
WHERE NOT EXISTS(SELECT *
FROM (SELECT A.P#
FROM SPJ AS A
WHERE A.S#='S1' ) AS C --注意集合里的元素的零件，零件与零件之间相比较
WHERE C.P# NOT IN(SELECT B.P#
FROM SPJ AS B
WHERE B.S#=S.S#));
标签：零件,--,数据库,SPJ,课后,SQL,WHERE,S1,select 来源： https://blog.csdn.net/qq_43615422/article/details/115654544