基本表结构：
student(sno,sname,sage,ssex)学生表
course(cno,cname,tno) 课程表
sc(sno,cno,score) 成绩表
teacher(tno,tname) 教师表
1，查询课程1的成绩比课程2的成绩高的所有学生的学号
select a.sno from
(select sno,score from sc where cno=1) a,
(select sno,score from sc where cno=2) b
where a.score>b.score and a.sno=b.sno
2，查询平均成绩大于60分的同学的学号和平均成绩
select a.sno as "学号", avg(a.score) as "平均成绩"
from
(select sno,score from sc) a
group by sno having avg(a.score)>60
3，查询所有同学的学号、姓名、选课数、总成绩
select a.sno as 学号, b.sname as 姓名,
count(a.cno) as 选课数, sum(a.score) as 总成绩
from sc a, student b
where a.sno = b.sno
group by a.sno, b.sname
或者：
select student.sno as 学号, student.sname as 姓名,
count(sc.cno) as 选课数, sum(score) as 总成绩
from student left Outer join sc on student.sno = sc.sno
group by student.sno, sname
4，查询姓“张”的老师的个数
select count(distinct(tname)) from teacher where tname like '张%‘
或者：
select tname as "姓名", count(distinct(tname)) as "人数"
from teacher
where tname like'张%'
group by tname
5，查询没学过“张三”老师课的同学的学号、姓名
select student.sno,student.sname from student
where sno not in (select distinct(sc.sno) from sc,course,teacher
where sc.cno=course.cno and teacher.tno=course.tno and teacher.tname='张三')
6，查询同时学过课程1和课程2的同学的学号、姓名
select sno, sname from student
where sno in (select sno from sc where sc.cno = 1)
and sno in (select sno from sc where sc.cno = 2)
或者：
selectc.sno, c.sname from
(select sno from sc where sc.cno = 1) a,
(select sno from sc where sc.cno = 2) b,
student c
where a.sno = b.sno and a.sno = c.sno
或者：
select student.sno,student.sname from student,sc where
student.sno=sc.sno and sc.cno=1
and exists( select * from sc as sc_2 where sc_2.sno=sc.sno and sc_2.cno=2)
7，查询学过“李四”老师所教所有课程的所有同学的学号、姓名
select a.sno, a.sname from student a, sc b
where a.sno = b.sno and b.cno in
(select c.cno from course c, teacher d where c.tno = d.tno and d.tname = '李四')
或者：
select a.sno, a.sname from student a, sc b,
(select c.cno from course c, teacher d where c.tno = d.tno and d.tname = '李四') e
where a.sno = b.sno and b.cno = e.cno
8，查询课程编号1的成绩比课程编号2的成绩高的所有同学的学号、姓名
select a.sno, a.sname from student a,
(select sno, score from sc where cno = 1) b,
(select sno, score from sc where cno = 2) c
where b.score > c.score and b.sno = c.sno and a.sno = b.sno
9，查询所有课程成绩小于60分的同学的学号、姓名
select sno,sname from student
where sno not in (select distinct sno from sc where score > 60)
10，查询至少有一门课程与学号为1的同学所学课程相同的同学的学号和姓名
select distinct a.sno, a.sname
from student a, sc b
where a.sno <> 1 and a.sno=b.sno and
b.cno in (select cno from sc where sno = 1)
或者：
select s.sno,s.sname
from student s,
(select sc.sno
from sc
where sc.cno in (select sc1.cno from sc sc1 where sc1.sno=1)and sc.sno<>1
group by sc.sno)r1
where r1.sno=s.sno
11. 请列出所有超过或等于5名学生的课
有一个courses 表 ，有: student (学生) 和 class (课程)。
请列出所有超过或等于5名学生的课。
例如,表:
±--------±-----------+
| student | class |
±--------±-----------+
| A | Math |
| B | English |
| C | Math |
| D | Biology |
| E | Math |
| F | Computer |
| G | Math |
| H | Math |
| I | Math |
±--------±-----------+
应该输出:
±--------+
| class |
±--------+
| Math |
±--------+
Note:
学生在每个课中不应被重复计算。
SELECT class FROM courses GROUP BY class having count(DISTINCT student)>=5
12. 获取 Employee 表中第 n 高的薪水（Salary）
编写一个 SQL 查询，获取 Employee 表中第 n 高的薪水（Salary）。
±—±-------+
| Id | Salary |
±—±-------+
| 1 | 100 |
| 2 | 200 |
| 3 | 300 |
±—±-------+
例如上述 Employee 表，n = 2 时，应返回第二高的薪水 200。如果不存在第 n 高的薪水，那么查询应返回 null。
±-----------------------+
| getNthHighestSalary(2) |
±-----------------------+
| 200 |
±-----------------------+
CREATE FUNCTION getNthHighestSalary(@N INT) RETURNS INT AS
BEGIN
Declare M INT;
Set M = N-1;
RETURN (
/* Write your T-SQL query statement below. */
select DISTINCT Salary from Employee order by Salary DESC limit M, 1
);
END
13. 编写一个 SQL 查询来实现分数排名
编写一个 SQL 查询来实现分数排名。如果两个分数相同，则两个分数排名（Rank）相同。请注意，平分后的下一个名次应该是下一个连续的整数值。换句话说，名次之间不应该有“间隔”。
±—±------+
| Id | Score |
±—±------+
| 1 | 3.50 |
| 2 | 3.65 |
| 3 | 4.00 |
| 4 | 3.85 |
| 5 | 4.00 |
| 6 | 3.65 |
±—±------+
例如，根据上述给定的 Scores 表，你的查询应该返回（按分数从高到低排列）：
±------±-----+
| Score | Rank |
±------±-----+
| 4.00 | 1 |
| 4.00 | 1 |
| 3.85 | 2 |
| 3.65 | 3 |
| 3.65 | 3 |
| 3.50 | 4 |
±------±-----+
SELECT Score,
(SELECT count(DISTINCT score) FROM Scores WHERE score >= s.score) AS Rank
FROM Scores s ORDER BY Score DESC ;