3 table S(student), C(course), SC(StudentCourse):
s(sno,sname)
c(cno,cname,cteacher)
sc(sno, cno, scgrade)
Q:
- 找出没选过“liming”老师的所有学生姓名.
- 列出2门以上（含2门）不及格学生姓名及平均成绩.
- 既学过1号课程又学过2号课所有学生的姓名.
A:
1,
1 select sname from s join sc on(s.sno = sc.sno) join c on(sc.cno = c.cno) 2 where cteacher <> ‘liming‘
2.1,
1 select sname from s where sno in 2 ( 3 select sno from sc where scgrade < 60 group by sno having count(*) >= 2 4 )
2.2,
1 select avg(scgrade) from sc join s on(sc.sno = s.sno) 2 where sname in 3 ( 4 select sname from s where sno in 5 ( 6 select sno from sc where scgrade < 60 group by sno having count(*) >= 2 7 ) 8 ) group by sname
3,
1 select sname from s join sc on(s.sno = sc.sno) 2 where sno in 3 ( 4 select sno from sc where cno = 1 and sno in 5 (select sno from sc where cno = 2) 6 )
ref: oracle