选读SQL经典实例笔记16_逻辑否定

1. 示例数据

1.1. student

insert into student values (1,'AARON',20)
insert into student values (2,'CHUCK',21)
insert into student values (3,'DOUG',20)
insert into student values (4,'MAGGIE',19)
insert into student values (5,'STEVE',22)
insert into student values (6,'JING',18)
insert into student values (7,'BRIAN',21)
insert into student values (8,'KAY',20)
insert into student values (9,'GILLIAN',20)
insert into student values (10,'CHAD',21)

1.2. courses

insert into courses values ('CS112','PHYSICS',4)
insert into courses values ('CS113','CALCULUS',4)
insert into courses values ('CS114','HISTORY',4)

1.3. professor

insert into professor values ('CHOI','SCIENCE',400,45)
insert into professor values ('GUNN','HISTORY',300,60)
insert into professor values ('MAYER','MATH',400,55)
insert into professor values ('POMEL','SCIENCE',500,65)
insert into professor values ('FEUER','MATH',400,40)

1.4. take

insert into take values (1,'CS112')
insert into take values (1,'CS113')
insert into take values (1,'CS114')
insert into take values (2,'CS112')
insert into take values (3,'CS112')
insert into take values (3,'CS114')
insert into take values (4,'CS112')
insert into take values (4,'CS113')
insert into take values (5,'CS113')
insert into take values (6,'CS113')
insert into take values (6,'CS114')

1.5. teach

insert into teach values ('CHOI','CS112')
insert into teach values ('CHOI','CS113')
insert into teach values ('CHOI','CS114')
insert into teach values ('POMEL','CS113')
insert into teach values ('MAYER','CS112')
insert into teach values ('MAYER','CS114')

2. 问题1：没有选修过CS112课程的学生

2.1. sql

select *from studentwhere sno in ( select snofrom takewhere cno != 'CS112' )

2.2. sql

select *from studentwhere sno not in (select snofrom takewhere cno = 'CS112')

2.3. 要记住真正的逻辑否定要求两个步骤，即为了找出‘哪些人不是’，就要先找出‘哪些人是’，然后再排除掉他们

2.4. PostgreSQL

2.5. MySQL

2.6. 使用CASE表达式和聚合函数MAX标识一个学生是否选修了CS112课程

2.6.1. sql

select s.sno,s.sname,s.agefrom student s left join take ton (s.sno = t.sno)group by s.sno,s.sname,s.age
having max(case when t.cno = 'CS112'then 1 else 0 end) = 0

2.7. Oracle

2.7.1. group by解决方案

select s.sno,s.sname,s.agefrom student s, take twhere s.sno = t.sno (+)group by s.sno,s.sname,s.agehaving max(case when t.cno = 'CS112'then 1 else 0 end) = 0

2.7.2. 窗口函数解决方案

select distinct sno,sname,agefrom (select s.sno,s.sname,s.age,max(case when t.cno = 'CS112'then 1 else 0 end)over(partition by s.sno,s.sname,s.age) as takes_CS112from student s, take twhere s.sno = t.sno (+)) xwhere takes_CS112 = 0

2.8. DB2

2.9. SQL Server

2.10. 使用CASE表达式和窗口函数MAX OVER

2.10.1. sql

select distinct sno,sname,agefrom (
select s.sno,s.sname,s.age,max(case when t.cno = 'CS112'then 1 else 0 end)over(partition by s.sno,s.sname,s.age) as takes_CS112from student s, take ton (s.sno = t.sno)) xwhere takes_CS112 = 0

2.11. 外连接到TAKE表是为了确保把那些没有选修任何课程的学生也能被筛选出来

2.12. 调用MAX函数找出最大的CASE表达式返回值

3. 问题2：只选修了CS112和CS114中的一门，而不是两门都选的学生

3.1. sql

select *from studentwhere sno in ( select snofrom takewhere cno != 'CS112'and cno != 'CS114' )

3.2. sql

select *from student s, take twhere s.sno = t.snoand t.cno in ( 'CS112', 'CS114' )and s.sno not in ( select a.snofrom take a, take bwhere a.sno = b.snoand a.cno = 'CS112'and b.cno = 'CS114' )

3.3. 使用自连接找出同时选修了CS112和CS114的学生

3.4. 使用子查询从选修了CS112或CS114的学生中把同时选了两门的学生剔除掉

3.5. DB2

3.6. Oracle

3.7. SQL Server

3.8. CASE表达式和窗口函数SUM OVER

3.8.1. sql

select distinct sno,sname,agefrom (
select s.sno,s.sname,s.age,sum(case when t.cno in ('CS112','CS114') then 1 else 0 end)over (partition by s.sno,s.sname,s.age) as takes_either_orfrom student s, take twhere s.sno = t.sno)xwhere takes_either_or = 1

3.9. PostgreSQL

3.10. MySQL

3.11. CASE表达式和聚合函数SUM

3.11.1. sql

select s.sno,s.sname,s.agefrom student s, take twhere s.sno = t.snogroup by s.sno,s.sname,s.age
having sum(case when t.cno in ('CS112','CS114')then 1 else 0 end) = 1

3.12. 内连接STUDENT表和TAKE表，这样就排除了那些没有选修任何课程的学生

3.13. 使用CASE表达式标记一个学生是否选修了这两门课程中的一门

3.14. 函数SUM会把每个学生对应的1都累加起来

4. 问题3:选修了CS112，而且没有选修其他课程的学生

4.1. sql

select s.*from student s, take twhere s.sno = t.snoand t.cno = 'CS112'

4.2. sql

select s.*from student s, take twhere s.sno = t.snoand s.sno not in ( select snofrom takewhere cno != 'CS112' )

4.3. 子查询负责找出至少选修了一门课，但又没有选修CS112的所有学生

4.4. 外层查询负责找出选修了一门课程（任意课程），并且不在上述子查询的返回结果的学生

4.5. STUDENT表和TAKE表之间的连接操作过滤掉没有选修任何课程的学生

4.6. PostgreSQL

4.7. MySQL

4.8. 使用聚合函数COUNT确保下列查询返回的学生只选修了一门课程

4.8.1. sql

select s.*from student s,take t1,(
select snofrom takegroup by sno
having count(*) = 1) t2where s.sno  = t1.snoand t1.sno = t2.snoand t1.cno = 'CS112'

4.8.2. 使用内嵌视图T2找出只选修了一门课程的学生

4.8.3. 连接内嵌视图T2到TAKE表，并且筛选出选修CS112课程的学生

4.8.4. 在内嵌视图T2和TAKE表连接查询的基础上再次连接STUDENT表，找出匹配的学生

4.9. DB2

4.10. Oracle

4.11. SQL Server

4.12. 使用窗口函数COUNT OVER

4.12.1. sql

select sno,sname,agefrom (
select s.sno,s.sname,s.age,t.cno,count(t.cno) over (partition by s.sno,s.sname,s.age) as cntfrom student s, take twhere s.sno = t.sno) xwhere cnt = 1and cno = 'CS112'