学生课程成绩

Student(Sid,Sname,Sage,Ssex) 学生表
Course(Cid,Cname,Tid) 课程表
SC(Sid,Cid,score) 成绩表
Teacher(Tid,Tname) 教师表 

1、查询“001”课程比“002”课程成绩高的所有学生的学号；

select a.sid from (select sid,score from sc where cid='001') a 
join (select sid,socre from sc    where cid='002') b on a.sid = b.sid 
where a.score > b.score;

2  查询平均成绩大于60分的同学的学号和平均成绩；

select sid,avg(score) from sc group by sid having avg(score)>60;

3  查询所有同学的学号、姓名、选课数、总成绩；

select stu.sid,stu.sname,count(cid),sum(score) from student stu join sc s on stu.sid = s.sid 

4 查询姓“李”的老师的个数；

select count(tid) from teacher where tname like '李%';

5 查询没学过“叶平”老师课的同学的学号、姓名；

  --叶平老师所带的课程编号
  select cid from course where tid = (select tid from teacher where tname="叶平");
  ---选过叶平老师课程的学生编号(重要)
  select distinct sid from sc where cid in (select cid from course where tid = (select tid from teacher where tname="叶平"));
  ---没有选过叶平老师课程的学生
  select sid,sname from student where sid not in (select distinct sid from sc where cid in (select cid from course where tid = (select tid from teacher where tname="叶平")));

6 查询学过“001”并且也学过编号“002”课程的同学的学号、姓名；

  --学过“001”---
  select sid from sc cid ='001';
  --学过“002”---
  select sid from sc cid ='002';
  --学过'001'同时学过'002'(重要 intersect)---
  select sid,sname from student where sid in ((select sid from sc cid ='001') intersect (select sid from sc cid ='002'));
  --- exists

用一条 SQL  语句查询出每门课都大于 80 

准备数据的 sql 代码：
create table score(id int primary key auto_increment,namevarchar(20),subject
varchar(20),score int);
insert into score values
(null,'张三','语文',81),
(null,'张三','数学',75),
(null,'李四','语文',76),
(null,'李四','数学',90),
(null,'王五','语文',81),
(null,'王五','数学',100),
(null,'王五 ','英语',90);
答案:
select distinct(name) from score where name not in (select distinct(name) from score where score <80)
或者
select distinct(name) from score t1 where 80 < all(select score from score where name = t1.name)

所有部门之间的比赛组合
一个叫 department 的表，里面只有一个字段 name,一共有 4
条纪录，分别是 a,b,c,d,对应四个球对，现在四个球对进行比赛，用一
条 sql 语句显示所有可能的比赛组合.

select a.name,b.name from team a, team b where a.name < b.name