1查询出只选修了一门课程的全部学生的学号和姓名
最初想法是select count(*)然后再去比较count(*),要是必就不会了
select s_id ,sname from student,sc where sc.s_id=student.id group by sc.s_id,sname having count(*)=1;
2查询同名同性学生名单，并统计同名人数
开始是让sc的2个表来比较同名的，然后发现错了多比了几遍，然后又想把id也相同。但是错的更离谱
select sname count(sname) from student group by id having count(sname)>1
3查询每门课程的平均成绩，结果按平均成绩升序排列，平均成绩相同时，按课程号降序排列
Select C#,Avg(score) from SC group by C# order by Avg(score),C# DESC ;
4查询平均成绩大于85的所有学生的学号、姓名和平均成绩
本意是用
select s_id from sc group by s_id having avg(score)>85;
有的s_id,然后再
select sname ,s_id avg(score) from sc,student where sc.s_id=student.id and s_id in (select avg(score),s_id from sc group by s_id having avg(score)>65) group by student.id;
但是sc.s_id=student.id and s_id ，结果
select sname,sc.s_id,avg(score) from sc,student where sc.s_id=student.id group by s_id,sname having avge(score) >85;
原文：http://www.cnblogs.com/bashala/p/3604351.html