SQL语句练习

前提条件

1.mysql环境,可以装个phpstudy,简单方便
2.建立数据表：
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
课程表：
Course(c_id,c_name,t_id) – –课程编号, 课程名称, 教师编号
教师表：
Teacher(t_id,t_name) –教师编号,教师姓名
成绩表：
Score(s_id,c_id,s_s_score) –学生编号,课程编号,分数

--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`)
);

3.填充数据

--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

1.查询课程编号为“01”的课程比“02”的课程成绩高的所有学生的学号

select a.*,b.s_score from
(select s_id,s_score from score where c_id='01') as a,
(select s_id,s_score from score where c_id = '02') as b
where a.s_score>b.s_score and a.s_id = b.s_id;

2.查询平均成绩大于60分的学生的学号和平均成绩

select s_id, avg(s_score)
from  score
group by s_id
having avg(s_score)>60;

3.查询所有学生的学号、姓名、选课数、总成绩

select st.s_id,st.s_name,count(sc.c_id),avg(sc.s_score)
from student st,score sc
where st.s_id = sc.s_id
group by s_id,s_name;

或者使用：

select st.s_id,st.s_name,count(sc.c_id),avg(sc.s_score)
from student st join score sc
on st.s_id = sc.s_id
group by s_id,s_name;

4.查询姓“张”的老师的个数

select count(distinct(t_name)) number from teacher where t_name like '张%';

5.查询没学过“张三”老师课的学生的学号、姓名

select distinct(st.s_id),st.s_name
from student st
where st.s_id not in
(select sc.s_id from score sc,teacher te,course co
where sc.c_id=co.c_id and te.t_id = co.t_id and t_name='张三');

6.查询学过“张三”老师所教的所有课的同学的学号、姓名

select distinct(st.s_id),st.s_name
from student st
where st.s_id in
(select sc.s_id from score sc,teacher te,course co
where sc.c_id=co.c_id and te.t_id = co.t_id and t_name='张三');

7.查询学过编号为“01”的课程并且也学过编号为“02”的课程的学生的学号、姓名

select distinct(st.s_id),st.s_name
from student st
where st.s_id in
(select a.s_id from (select s_id,c_id from score where c_id='01') a
inner join (select s_id,c_id  from score where c_id='02') b
on a.s_id=b.s_id
);

或者（下面这个可以很好地改为学过“01”没学过“02”等情况）

select distinct(st.s_id),st.s_name
from student st
where st.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')

8.查询课程编号为“02”的总成绩

select sum(s_score)
from score
where c_id='02';

9.查询存在课程成绩小于60分的学生的学号、姓名

select distinct(st.s_id), st.s_name
from student st,score sc
where st.s_id=sc.s_id and sc.s_score<60;

如果是所有课程小于60分，则为

select distinct(st.s_id), st.s_name
from student st
where st.s_id not in
(select s_id from score where s_score>60)

10.查询没有学全所有课的学生的学号、姓名

select st.s_id,st.s_name
from student st inner join score sc on st.s_id=sc.s_id
group by st.s_id,st.s_name
having count(c_id)<(select count(distinct(c_id)) from course);

11.查询至少有一门课与学号为“01”的学生所学课程相同的学生的学号和姓名

select distinct(st.s_id),st.s_name
from student st inner join score sc on st.s_id=sc.s_id
where sc.c_id in
(select c_id from score where s_id='01') and st.s_id !='01';

12.查询和“01”号同学所学课程完全相同的其他同学的学号

select sc.s_id, st.s_name
from score sc inner join student st on sc.s_id=st.s_id
group by sc.s_id,st.s_name
having group_concat(c_id order by c_id)
=(select group_concat(c_id order by c_id) from score where s_id='01')
and sc.s_id!='01'

13.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 s_id not in
(select sc.s_id
from score sc
inner join course co on sc.c_id=co.c_id
inner join teacher te on co.t_id=co.t_id
where te.t_name='张三');

或者

select distinct st.s_id,st.s_name
from student st
where st.s_id not in
(select sc.s_id from score sc,teacher te,course co
where co.t_id = te.t_id and sc.c_id=co.c_id and te.t_name='张三');

14.查询两门及其以上不及格课程的同学的学号，姓名及其平均成绩

select distinct st.s_id,st.s_name ,avg(sc.s_score)
from student st inner join score sc on st.s_id=sc.s_id
where not sc.s_score >= 60
group by st.s_id,st.s_name
having count(c_id)>=2;

或

SELECT student.s_id,student.s_name,avg(score.s_score)
From student inner join score on student.s_id = score.s_id
Where student.s_id in
(select tem.s_id from (select * from score sc where sc.s_score<60) tem
group by tem.s_id
having count(tem.c_id)>=2)
group by student.s_id,student.s_name

或

SELECT student.s_id,student.s_name,avg(score.s_score)
From student inner join score on student.s_id = score.s_id
Where student.s_id in
(select sc.s_id from score sc where sc.s_score<60
group by sc.s_id
having count(sc.c_id)>=2)
group by student.s_id,student.s_name

15.检索"01"课程分数小于60，按分数降序排列的学生信息

select *
from student st inner join score sc
on st.s_id=sc.s_id
where sc.s_score<60 and sc.c_id='01'
order by sc.s_score desc;

16.按平均成绩从高到低显示所有学生的所有课程的成绩以及平均成绩

select s_id,
max(case when c_id='01' then s_score else NUll end) '01',
max(case when c_id='02' then s_score else NULL end) '02',
max(case when c_id='03' then s_score else NUll end) '03',
avg(s_score)
from score
group by s_id
order by avg(s_score) desc

17.查询各科成绩最高分、最低分和平均分：以如下形式显示：课程ID，课程name，最高分，最低分，平均分，及格率，中等率，优良率，优秀率

及格为>=60，中等为：70-80，优良为：80-90，优秀为：>=90

select sc.c_id, co.c_name, max(s_score) max, min(s_score) min, avg(s_score) avg,
avg(case when s_score>=60 then 1.0 else 0.0 end) '及格率',
avg(case when s_score>=70 and s_score<80 then 1.0 else 0.0 end) '中等率',
avg(case when s_score>=80 and s_score<90 then 1.0 else 0.0 end) '优良率',
avg(case when s_score>=90 then 1.0 else 0.0 end) '优秀率'
from score sc inner join course co on sc.c_id=co.c_id
group by sc.c_id;

18.查询每门课程的平均成绩，结果按平均成绩升序排序，平均成绩相同时，按课程号降序排列

select c_id, avg(s_score)
from score
group by c_id
order by avg(s_score) asc,c_id desc;

19.查询学生的总成绩并进行排名

select s_id,sum(s_score)
from score sc
group by s_id
order by sum(s_score) desc;

20.查询不同老师所教不同课程平均分从高到低显示

select te.t_id, te.t_name, sc.c_id,avg(sc.s_score)
from score sc inner join course co on sc.c_id=co.c_id
inner join teacher te on co.t_id=te.t_id
group by te.t_id,te.t_name,sc.c_id
order by avg(sc.s_score) desc;

参考：

1.sql面试题（学生表_课程表_成绩表_教师表）
2.SQL面试必会50题