Mysql练习题

Posted 2022-12-31 liangyan131

表名和字段
------------------------------------------------------------------------------
–1.学生表
Student(s_id,s_name,s_birth,s_sex) --学生编号,学生姓名, 出生年月,学生性别
–2.课程表
Course(c_id,c_name,t_id) – --课程编号, 课程名称, 教师编号额
–3.教师表
Teacher(t_id,t_name) --教师编号,教师姓名
–4.成绩表
Score(s_id,c_id,s_score) --学生编号,课程编号,分数

测试数据
------------------------------------------------------------------------------
--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);
---------------------

练习题
------------------------------------------------------------------------------
-- 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
s1.*,
s2.01_score,
s2.02_score
from
student s1,
(
select t1.s_id as s_id,
t1.s_score as 01_score,
t2.s_score as 02_score
from
score t1,
score t2
where
t1.s_id = t2.s_id
and t1.c_id = ‘01‘
and t2.c_id = ‘02‘
and t1.s_score > t2.s_score ) s2
where
s1.s_id = s2.s_id;
-- 2、查询"01"课程比"02"课程成绩低的学生的信息及课程分数
select
s1.*,
s2.01_score,
s2.02_score
from
student s1,
(
select t1.s_id as s_id,
t1.s_score as 01_score,
t2.s_score as 02_score
from
score t1,
score t2
where
t1.s_id = t2.s_id
and t1.c_id = ‘01‘
and t2.c_id = ‘02‘
and t1.s_score < t2.s_score ) s2
where
s1.s_id = s2.s_id;
-- 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
t1.s_id,
t2.s_name,
avg(t1.s_score) as avg_score
from
score t1
left join student t2 on
t1.s_id = t2.s_id
group by
t1.s_id
having
avg(t1.s_score) >= 60;
-- 4、查询平均成绩小于60分的同学的学生编号和学生姓名和平均成绩
-- (包括有成绩的和无成绩的)
select
t1.s_id,
t2.s_name,
avg(t1.s_score) as avg_score
from
score t1
left join student t2 on
t1.s_id = t2.s_id
group by
t1.s_id
having
avg(t1.s_score) < 60;
-- 5、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
select
t2.s_id,
t2.s_name,
count(t1.c_id) as sun_course,
sum(t1.s_score) as sum_score
from
student t2
left join score t1 on
t1.s_id = t2.s_id
group by
t1.s_id;
-- 6、查询"李"姓老师的数量
select
count(1)
from
teacher
where
t_name like ‘李%‘ ;
-- 7、查询学过"张三"老师授课的同学的信息
select
t1.*
from
student t1,
score t2
where
t1.s_id = t2.s_id
and t2.c_id in (
select c_id
from
course
where
t_id = (
select t_id
from
teacher
where
t_name = ‘张三‘));
-- 8、查询没学过"张三"老师授课的同学的信息
select
*
from
student
where
s_id not in (
select
distinct s_id
from
score
where
c_id in (
select t2.c_id
from
teacher t1,
course t2
where
t1.t_id = t2.t_id
and t1.t_name = ‘张三‘));