接之前的SQL除法，我们来一次SQL除法的加强版。实际工作中这种需求可能很少，但是为应付面试题，没准有用。
1.数据脚本
drop table student;
drop table subject;
create table student(stu_name varchar2(20),subject_id varchar2(10));
create table subject(subject_id varchar2(10),subject_name varchar2(20),teacher varchar2(20));
insert into student values('狗蛋','M');
insert into student values('狗蛋','C');
insert into student values('狗蛋','P');
insert into student values('翠花','M');
insert into student values('翠花','C');
insert into student values('小红','M');
insert into student values('小红','P');
insert into student values('小美','M');
insert into student values('小美','T');
insert into student values('阿健','M');
insert into student values('阿健','C');
insert into student values('阿健','P');
insert into student values('阿健','T');
insert into subject values('M','数学','张老师');
insert into subject values('C','语文','张老师');
insert into subject values('M','数学','李老师');
insert into subject values('C','语文','李老师');
insert into subject values('P','物理','李老师');
insert into subject values('M','数学','孙老师');
insert into subject values('P','物理','孙老师');
insert into subject values('T','体育','何老师');
commit;
2. SQL需求
要求检索学生的选课来自同一位老师的学生信息和老师信息。
3. SQL实现
select distinct sd1.stu_name,sj1.teacher
from student sd1,
subject sj1
where not exists (select 1 --①
from student sd2 --②
where sd1.stu_name=sd2.stu_name
and not exists (select 1 --④
from subject sj2
where sd2.subject_id=sj2.subject_id
and sj1.teacher=sj2.teacher)) --③
order by 1,2;
--①，查询返回的条件是这一层not exists返回0条数据
--②，sd1和sd2用stu_name进行关联，以学生为单位，从sd2中得到学生的选课（subject_id）结果集
--③，用sj1和sj2的teacher进行关联，得到老师教课的结果集
--④，只有②的每一条subject_id都在这一层not exists检索的subject_id中，查询①处的结果才是0条，即要求②是③的子集
