题 目 ：
student(sid, sname, sage, ssex) -- 学生信息表(学生编号 自增，学生姓名， 学生出生年月， 性别)；
teacher(tid, tname) -- 教师信息表（教师编号 自增， 教师姓名）
course(cid, cname, tid) -- 课程表（课程编号 自增， 课程名称， 教师编号 外键）
sc(sid, cid, score) -- 分数表（学生编号，课程编号，分数）
create database work;
use work;
set names work;
# 创建表
create table student(
sid int primary key auto_increment,
sname varchar(20),
sage date,
ssex varchar(10)
)
create table teacher(
tid int primary key auto_increment,
tname varchar(20)
);
create table course(
cid int primary key auto_increment,
cname varchar(20),
tid int ,
foreign key(tid) references teacher(tid)
)
create table sc(
sid int,
cid int,
score int
)
# 插入数据
# 先给student表插入数据
insert into student values
(1,‘赵雷‘,‘1990-01-01‘,‘男‘),
(2,‘钱电‘,‘1990-12-21‘,‘男‘),
(3,‘孙风‘,‘1990-05-20‘,‘男‘),
(4,‘李云‘,‘1990-08-06‘,‘男‘),
(5,‘周梅‘,‘1991-12-01‘,‘女‘),
(6,‘吴兰‘,‘1992-03-01‘,‘女‘),
(7,‘郑竹‘,‘1989-07-01‘,‘女‘),
(8,‘王菊‘,‘1990-01-20‘,‘女‘);
# 教师表
insert into teacher values
(1,‘张三‘),
(2,‘李四‘),
(3,‘王五‘);
# 课程表
insert into course values
(1,‘语文‘,2),
(2,‘语文‘,1),
(3,‘语文‘,3);
# 分数表
insert into sc values
(1,1,90),
(1,2,80),
(1,3,90),
(2,1,70),
(2,2,60),
(2,3,80),
(3,1,80),
(3,2,80),
(3,3,80),
(4,1,50),
(4,2,30),
(4,3,20),
(5,1,76),
(5,2,87),
(6,1,31),
(6,3,34),
(7,2,89),
(7,3,98);
问 题：
-- 1、查询”01”课程比”02”课程成绩高的学生的信息及课程分数
select
st.sid,
st.sname,
st.sage,
st.ssex,
sc1.score as 课程1分数,
sc2.score as 课程1分数
from sc sc1,
sc sc2,
student st
where sc1.sid = sc2.sid
and sc1.sid = st.sid
and sc1.cid = 1
and sc2.cid = 2
and sc1.score > sc2.score
-- 2、查询学生表前5名信息；
select * from student st limit 5;
-- 3.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
s.sname,
sc.sid,
avg(sc.score)
from student s,
sc
where s.sid = sc.sid
group by sc.sid
having avg(sc.score) >= 60
-- 4、查询名字中含有”风”字的学生信息
select * from student st where st.sname like ‘%风‘;
-- 5、查询课程名称为”数学”,且分数低于60的学生姓名和分数
select
st.sname,
sc.score
from sc,
student st
where st.sid = sc.sid
and sc.score < 60
and sc.cid = (select
co.cid
from course co
where co.cname = ‘数学‘)
-- 6、查询所有学生的课程及分数情况;
select
st.sname,
co.cname,
sc.score
from student st,
sc,
course co
where st.sid = sc.sid
and co.cid = sc.cid;
-- 7、查询没学过”张三”老师授课的同学的信息
select *
from student st
where st.sid not in(select st1.sid
from student st1,
sc,
course co,
teacher te
where st1.sid = sc.sid
and co.cid = sc.cid
and co.tid = te.tid
and te.tname = ‘张三‘) ;
-- 8、查询学过”张三”老师授课的同学的信息
select
st1.*
from student st1,
sc,
course co,
teacher te
where st1.sid = sc.sid
and co.cid = sc.cid
and co.tid = te.tid
and te.tname = ‘张三‘
-- 9、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
st.sid,
st.sname,
st.sage,
st.ssex
from sc sc1,
sc sc2,
student st
where sc1.sid = sc2.sid
and sc1.sid = st.sid
and sc1.cid = 1
and sc2.cid = 2
-- 10、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
st.sid,
st.sname,
st.sage,
st.ssex
from student st,
sc
where st.sid = sc.sid
and sc.cid = 1
and st.sid not in(select distinct
sc1.sid
from sc sc1,
sc sc2
where sc1.cid = 2)
-- 11、查询没有学全所有课程的同学的信息
select * from student st
where st.sid not in
(select
distinct st.sid
from sc sc1,
sc sc2,
sc sc3,
student st
where sc1.sid = sc2.sid
and sc1.sid = sc3.sid
and sc1.sid = st.sid
and sc1.cid = 1
and sc2.cid = 2
and sc3.cid = 3)
-- 12、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,
sc
where st.sid = sc.sid
and sc.cid in(select
sc.cid
from sc
where sc.sid = 1)
and st.sid != 1
-- 13、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.sid, st.sname, st.sage, st.ssex
from student st,
sc
where st.sid = sc.sid
and st.sid != 1
and sc.cid in(select
sc.cid
from sc
where sc.sid = 1)
group by st.sid
having count( * ) = (select
count(*)
from sc
where sc.sid = 1)
-- 14、查询没学过”张三”老师讲授的任一门课程的学生姓名
#方法1：
select *
from student st
where st.sid not in(select
st.sid
from student st,
sc
where sc.sid = st.sid
and sc.cid = (select
co.cid
from course co
where co.tid = (select
te.tid
from teacher te
where te.tname = ‘张三‘)))
#方法2：
select
s.*
from student s
where s.sid not in(select
sc1.sid
from sc sc1,
course c,
teacher t
where sc1.cid = c.cid
and c.tid = t.tid
and t.tname = ‘张三‘);
-- 15、查询出只有两门课程的全部学生的学号和姓名
select
st.sid,
st.sname,
st.sage,
st.ssex
from student st,
sc
where st.sid = sc.sid
group by sc.sid
having count( * ) = 2
-- 16、查询1990年出生的学生名单(注:Student表中Sage列的类型是datetime)
select * from student st where st.sage >= ‘1990-01-01‘ and st.sage <= ‘1990-12-31‘;
select s.* from student s where s.sage like ‘1990-%‘;
select * from student st where st.sage between ‘1990-01-01‘ and ‘1990-12-31‘;
-- 17、查询每门课程的平均成绩,结果按平均成绩降序排列,平均成绩相同时,按课程编号升序排列
select sc.cid, avg(sc.score) from sc group by sc.cid order by avg(sc.score) desc, sc.cid
-- 18、查询任何一门课程成绩在70分以上的姓名、课程名称和分数;
#方法1：
select
s.sname,
c.cname,
score
from student s,
sc,
course c
where s.sid = sc.sid
and sc.cid = c.cid
and score > 70;
#方法2：
select st.sname, co.cname, sc.score
from student st,
course co,
sc
where st.sid = sc.sid
and sc.cid = co.cid
and st.sid in(select
sc.sid
from sc
group by sc.sid
having min(sc.score) >= 70)
-- 19、查询平均成绩大于等于85的所有学生的学号、姓名和平均成绩
select
st.sid,
st.sname,
avg(sc.score) avgscore
from student st,
sc
where st.sid = sc.sid
group by sc.sid
having avg(sc.score) >= 85
-- 20、查询不及格的课程
select
st.sname,
co.cname,
sc.score
from student st,
course co,
sc
where st.sid = sc.sid
and sc.cid = co.tid
and sc.score < 60
-- 21、查询课程编号为01且课程成绩在80分以上的学生的学号和姓名;
select
st.sid,
st.sname
from student st,
sc
where st.sid = sc.sid
and sc.cid = 1
and sc.score < 60
-- 22、求每门课程的学生人数
select sc.cid, count(*) from sc group by sc.cid
select cid,count(sid) from sc group by sc.cid;
-- 23、统计每门课程的学生选修人数(超过5人的课程才统计)。要求输出课程号和选修人数,查询结果按人数降序排列,若人数相同,按课程号升序排列
select
sc.cid,
count(*) sumstudent
from sc
group by sc.cid
having count( * ) > 5
order by sumstudent desc, sc.cid
-- 24、查询不同课程成绩相同的学生的学生编号、课程编号、学生成绩
select
sc1.sid, sc1.cid,sc2.sid, sc2.cid,sc1.score
from sc sc1,
sc sc2
where sc1.sid != sc2.sid
and sc1.cid != sc2.cid
and sc1.score = sc2.score
-- 25、检索至少选修两门课程的学生学号
select
sc.sid
from sc
group by sc.sid
having count( * ) >= 2
-- 26、查询选修了全部课程的学生信息
select
st.*
from sc,
student st
where st.sid = sc.sid
group by sid
having count(sc.cid) = 3;
-- 27、查询各学生的年龄
select s.sname,(to_days(‘2018-06-23‘)-to_days(s.sage))/365 as age from student s;
-- 28、查询本月过生日的学生
select s.sname from student s where s.sage like ‘_____07%‘;
-- 39、查询学全所有课程的同学的信息
select
s.*
from student s,
sc sc1,
sc sc2,
sc sc3
where sc1.cid = 1
and sc2.cid = 2
and sc3.cid = 3
and sc1.sid = sc2.sid
and sc1.sid = sc3.cid
and s.sid = sc1.sid
group by s.sid;
-- 30、查询课程2 第2名到第5名的分数，降序排列
select * from course co, sc where co.cid = sc.cid and co.cid=2 order by sc.score desc limit 1,4
原文地址：https://www.cnblogs.com/albatron/p/9218778.html