建表：
-- 建表
-- 学生表
CREATE TABLE `Student`(
`s_id` VARCHAR(20),
`s_name` VARCHAR(20) NOT NULL DEFAULT '',
`s_birth` VARCHAR(20) NOT NULL DEFAULT '',
`s_sex` VARCHAR(10) NOT NULL DEFAULT '',
PRIMARY KEY(`s_id`)
);
-- 课程表
CREATE TABLE `Course`(
`c_id` VARCHAR(20),
`c_name` VARCHAR(20) NOT NULL DEFAULT '',
`t_id` VARCHAR(20) NOT NULL,
PRIMARY KEY(`c_id`)
);
-- 教师表
CREATE TABLE `Teacher`(
`t_id` VARCHAR(20),
`t_name` VARCHAR(20) NOT NULL DEFAULT '',
PRIMARY KEY(`t_id`)
);
-- 成绩表
CREATE TABLE `Score`(
`s_id` VARCHAR(20),
`c_id` VARCHAR(20),
`s_score` INT(3),
PRIMARY KEY(`s_id`,`c_id`)
);
-- 插入学生表测试数据
insert into Student values('01' , '赵雷' , '1990-01-01' , '男');
insert into Student values('02' , '钱电' , '1990-12-21' , '男');
insert into Student values('03' , '孙风' , '1990-05-20' , '男');
insert into Student values('04' , '李云' , '1990-08-06' , '男');
insert into Student values('05' , '周梅' , '1991-12-01' , '女');
insert into Student values('06' , '吴兰' , '1992-03-01' , '女');
insert into Student values('07' , '郑竹' , '1989-07-01' , '女');
insert into Student values('08' , '王菊' , '1990-01-20' , '女');
-- 课程表测试数据
insert into Course values('01' , '语文' , '02');
insert into Course values('02' , '数学' , '01');
insert into Course values('03' , '英语' , '03');
-- 教师表测试数据
insert into Teacher values('01' , '张三');
insert into Teacher values('02' , '李四');
insert into Teacher values('03' , '王五');
-- 成绩表测试数据
insert into Score values('01' , '01' , 80);
insert into Score values('01' , '02' , 90);
insert into Score values('01' , '03' , 99);
insert into Score values('02' , '01' , 70);
insert into Score values('02' , '02' , 60);
insert into Score values('02' , '03' , 80);
insert into Score values('03' , '01' , 80);
insert into Score values('03' , '02' , 80);
insert into Score values('03' , '03' , 80);
insert into Score values('04' , '01' , 50);
insert into Score values('04' , '02' , 30);
insert into Score values('04' , '03' , 20);
insert into Score values('05' , '01' , 76);
insert into Score values('05' , '02' , 87);
insert into Score values('06' , '01' , 31);
insert into Score values('06' , '03' , 34);
insert into Score values('07' , '02' , 89);
insert into Score values('07' , '03' , 98);
*题目一：查询" 01 "课程比" 02 "课程成绩高的学生的信息及课程分数
select n1.s_id, score1, score2
from (select s_id, s_score as score1 from score where c_id=01) n1 join (select s_id, s_score as score2 from score where c_id=02) n2 on n1.s_id=n2.s_id
where score1>score2;
1.1 查询同时存在" 01 "课程和" 02 "课程的情况
select n1.s_id, score1, score2
from (select s_id, s_score as score1 from score where c_id=01) n1 join (select s_id, s_score as score2 from score where c_id=02) n2 on n1.s_id=n2.s_id;
1.2 查询存在" 01 "课程但可能不存在" 02 "课程的情况(不存在时显示为 null )
select n1.s_id, score1, score2
from (select s_id, s_score as score1 from score where c_id=01) n1 left join (select s_id, s_score as score2 from score where c_id=02) n2 on n1.s_id=n2.s_id;
1.3 查询不存在" 01 "课程但存在" 02 "课程的情况
select n2.s_id, score1, score2
from (select s_id, s_score as score1 from score where c_id=01) n1 right join (select s_id, s_score as score2 from score where c_id=02) n2 on n1.s_id=n2.s_id;
题目二：查询平均成绩大于等于 60 分的同学的学生编号和学生姓名和平均成绩
select st.s_id, st.s_name, avg(s_score) as avg_score
from student st join score sc on st.s_id=sc.s_id
group by st.s_id
having avg_score >= 60;
题目三：查询在 SC 表存在成绩的学生信息
select distinct st.s_id, st.s_name, avg(sc.s_score)
from student st join score sc on st.s_id=sc.s_id
group by st.s_id;
题目四：查询所有同学的学生编号、学生姓名、选课总数、所有课程的总成绩(没成绩的显示为null)
select st.s_id, st.s_name, count(s_score) as num_course, sum(s_score) as sum_score
from student st left join score sc on st.s_id=sc.s_id
group by st.s_id;
4.1 查有成绩的学生信息
select st.s_id, st.s_name, count(s_score) as num_course, sum(s_score) as sum_score
from student st join score sc on st.s_id=sc.s_id
group by st.s_id;
题目五：查询「李」姓老师的数量
select count(t_name) as num_teacher
from teacher
where t_name like '李%';
（或者使用regexp '李.'）
题目六：查询学过「张三」老师授课的同学的信息
select a.s_id, a.s_name,a.s_birth, a.s_sex, a.c_id, b.t_name
from (select st.s_id, st.s_birth, st.s_sex, s_name, c_id from student st join score sc on st.s_id=sc.s_id) a
join (select co.c_id, te.t_name from teacher te join course co on te.t_id=co.t_id) b on a.c_id=b.c_id
where b.t_name='张三';
经过与他人答案对比，有以下更简单的方法：
select student.*
from student, score sc
where student.s_id = sc.s_id
and sc.c_id in (select c_id from course, teacher where course.t_id = teacher.t_id and T_name = '张三')
*6.1查询学过编号为"01"并且也学过编号为"02"的课程的同学的信息
select * from student
where s_id in (select s_id from score where c_id=01)
and s_id in (select s_id from score where c_id=02);
6.2查询学过编号为"01"但是没有学过编号为"02"的课程的同学的信息
select * from student
where s_id in (select s_id from score where c_id=01)
and s_id not in (select s_id from score where c_id=02);
题目七：查询没有学全所有课程的同学的信息
select * from student
where s_id not in (select s_id from score where c_id=01)
or s_id not in (select s_id from score where c_id=02)
or s_id not in (select s_id from score where c_id=03);
（也可以使用count查询学习课程数<总课程数的学生）
题目八：查询至少有一门课与学号为" 01 "的同学所学相同的同学的信息
select distinct st.* from student st, score sc
where st.s_id=sc.s_id and sc.c_id in (select c_id from score where s_id=01);
***题目九：查询和" 01 "号的同学学习的课程 完全相同的其他同学的信息
select st.* from student st where st.s_id in
(select distinct s_id from score where s_id!='01' and c_id in(select c_id from score where s_id='01')
group by s_id
having count(c_id)=(select count(c_id) from score where s_id='01'));
题目十：查询没学过"张三"老师讲授的任一门课程的学生姓名
select st.s_id, st.s_name
from student st
where st.s_id not in (select s_id
from score sc where sc.c_id in (select c_id from course co, teacher te where co.t_id=te.t_id and te.t_name='张三'));
（思路:先找出学过张三老师任一门课程的同学，将这些同学排除即可）