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 student where sno in ( select sno                  from take                 where cno != 'CS112' )

2.2. sql

select *  from student where sno not in (select sno                     from take                    where 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. PostgreSQL2.5. MySQL2.6. 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.age  from student s left join take t    on (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. Oracle2.7.1. group by解决方案

select s.sno,s.sname,s.age   from student s, take t  where 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

2.7.2. 窗口函数解决方案

select distinct sno,sname,age   from ( 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_CS112   from student s, take t  where s.sno = t.sno (+)        ) x  where takes_CS112 = 0

2.8. DB22.9. SQL Server2.10. 使用CASE表达式和窗口函数MAX OVER2.10.1. sql

select distinct sno,sname,age  from (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_CS112  from student s, take t    on (s.sno = t.sno)       ) x where 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 student where sno in ( select sno                  from take                 where cno != 'CS112'                   and cno != 'CS114' )

3.2. sql

select *  from student s, take t where s.sno = t.sno   and t.cno in ( 'CS112', 'CS114' )   and s.sno not in ( select a.sno                      from take a, take b                     where a.sno = b.sno                       and 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. DB23.6. Oracle3.7. SQL Server3.8. CASE表达式和窗口函数SUM OVER3.8.1. sql

select distinct sno,sname,age  from (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_or  from student s, take t where s.sno = t.sno       )x where takes_either_or = 1

3.9. PostgreSQL3.10. MySQL3.11. CASE表达式和聚合函数SUM3.11.1. sql

select s.sno,s.sname,s.age  from student s, take t where s.sno = t.sno group by s.sno,s.sname,s.agehaving 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 t where s.sno = t.sno   and t.cno = 'CS112'

4.2. sql

select s.*  from student s, take t where s.sno = t.sno   and s.sno not in ( select sno                        from take                       where 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. PostgreSQL4.7. MySQL4.8. 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 sno   from take group by snohaving count(*) = 1       ) t2 where s.sno  = t1.sno   and t1.sno = t2.sno   and 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. DB24.10. Oracle4.11. SQL Server4.12. 使用窗口函数COUNT OVER4.12.1. sql

select sno,sname,age  from (select s.sno,s.sname,s.age,t.cno,       count(t.cno) over (         partition by s.sno,s.sname,s.age       ) as cnt  from student s, take t where s.sno = t.sno       ) x where cnt = 1   and cno = 'CS112'

4.12.2. 窗口函数解决方案处理方式上稍有不同（更有效率）4.12.3. 内嵌视图X返回了每一个学生、他们选修的课程以及他们选修了几门课程4.12.4. 获得了每个学生选修的课程和课程数目之后，最后只要保留CNT等于1并且CNO等于CS112的行即可