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MySQL 练习题（经典 50 题）
– 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)
);
– 老师表
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’);
– 老师表测试数据
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’ , ‘0’ , ’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’);
– 1、查问 ”01″ 课程比 ”02″ 课程问题高的学生的信息及课程分数
select st.*,sc.s_score as‘语文’,sc2.s_score‘数学’
from student st
left join score sc on sc.s_id=st.s_id and sc.c_id=‘01’
left join score sc2 on sc2.s_id=st.s_id and sc2.c_id=‘02’
where sc.s_score>sc2.s_score
– 2、查问 ”01″ 课程比 ”02″ 课程问题低的学生的信息及课程分数
select st.*,sc.s_score‘语文’,sc2.s_score‘数学’from student st
left join score sc on sc.s_id=st.s_id and sc.c_id=‘01’
left join score sc2 on sc2.s_id=st.s_id and sc2.c_id=‘02’
where sc.s_score<sc2.s_score
– 3、查问均匀问题大于等于 60 分的同学的学生编号和学生姓名和均匀问题
select st.s_id,st.s_name,ROUND(AVG(sc.s_score),2) cjScore from student st
left join score sc on sc.s_id=st.s_id
group by st.s_id having AVG(sc.s_score)>=60
– 4、查问均匀问题小于 60 分的同学的学生编号和学生姓名和均匀问题
– (包含有问题的和无问题的)
select st.s_id,st.s_name,(case when ROUND(AVG(sc.s_score),2) is null then 0 else ROUND(AVG(sc.s_score)) end ) cjScore from student st
left join score sc on sc.s_id=st.s_id
group by st.s_id having AVG(sc.s_score)<60 or AVG(sc.s_score) is NULL
– 5、查问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 st.s_id,st.s_name,count(c.c_id),(case when SUM(sc.s_score) is null or sum(sc.s_score)=“”then 0 else SUM(sc.s_score) end) from student st
left join score sc on sc.s_id =st.s_id
left join course c on c.c_id=sc.c_id
group by st.s_id
– 6、查问 ” 李 ” 姓老师的数量
select t.t_name,count(t.t_id) from teacher t
group by t.t_id having t.t_name like“李 %”;
– 7、查问学过 ” 张三 ” 老师授课的同学的信息
select st.* from student st
left join score sc on sc.s_id=st.s_id
left join course c on c.c_id=sc.c_id
left join teacher t on t.t_id=c.t_id
where t.t_name=“张三”
– 8、查问没学过 ” 张三 ” 老师授课的同学的信息
– 张三老师教的课
select c.* from course c left join teacher t on t.t_id=c.t_id where t.t_name=“张三”
– 有张三老师课问题的 st.s_id
select sc.s_id from score sc where sc.c_id in (select c.c_id from course c left join teacher t on t.t_id=c.t_id where t.t_name=“张三”)
– 不在下面查到的 st.s_id 的学生信息, 即没学过张三老师授课的同学信息
select st.* from student st where st.s_id not in(
select sc.s_id from score sc where sc.c_id in (select c.c_id from course c left join teacher t on t.t_id=c.t_id where t.t_name=“张三”)
)
– 9、查问学过编号为 ”01″ 并且也学过编号为 ”02″ 的课程的同学的信息
select st.* from student st
inner join score sc on sc.s_id = st.s_id
inner join course c on c.c_id=sc.c_id and c.c_id=“01”
where st.s_id in (
select st2.s_id from student st2
inner join score sc2 on sc2.s_id = st2.s_id
inner join course c2 on c2.c_id=sc2.c_id and c2.c_id=“02”
)
网友提供的思路(厉害呦~):
SELECT st.*
FROM student st
INNER JOIN score sc ON sc.s_id=st.s_id
GROUP BY st.s_id
HAVING SUM(IF(sc.c_id=“01”OR sc.c_id=“02”,1,0))>1
– 10、查问学过编号为 ”01″ 然而没有学过编号为 ”02″ 的课程的同学的信息
select st.* from student st
inner join score sc on sc.s_id = st.s_id
inner join course c on c.c_id=sc.c_id and c.c_id=“01”
where st.s_id not in (
select st2.s_id from student st2
inner join score sc2 on sc2.s_id = st2.s_id
inner join course c2 on c2.c_id=sc2.c_id and c2.c_id=“02”
)
– 11、查问没有学全所有课程的同学的信息
– 太简单, 下次换一种思路, 看有没有简略点办法
– 此处思路为查学全所有课程的学生 id, 再内联取背面
select * from student where s_id not in (
select st.s_id from student st
inner join score sc on sc.s_id = st.s_id and sc.c_id=“01”
where st.s_id in (
select st2.s_id from student st2
inner join score sc2 on sc2.s_id = st2.s_id and sc2.c_id=“02”
) and st.s_id in (
select st2.s_id from student st2
inner join score sc2 on sc2.s_id = st2.s_id and sc2.c_id=“03”
))
– 来自一楼网友的思路，左连贯，依据学生 id 分组过滤掉 数量小于 课程表中总课程数量的后果(show me his code), 简洁不少。
select st.* from Student st
left join Score S
on st.s_id = S.s_id
group by st.s_id
having count(c_id)<(select count(c_id) from Course)
– 12、查问至多有一门课与学号为 ”01″ 的同学所学雷同的同学的信息
select distinct st.* from student st
left join score sc on sc.s_id=st.s_id
where sc.c_id in (
select sc2.c_id from student st2
left join score sc2 on sc2.s_id=st2.s_id
where st2.s_id =‘01’
)
– 13、查问和 ”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
left join score sc on sc.s_id=st.s_id
group by st.s_id
having group_concat(sc.c_id) =
(
select group_concat(sc2.c_id) from student st2
left join score sc2 on sc2.s_id=st2.s_id
where st2.s_id =‘01’
)
– 14、查问没学过 ” 张三 ” 老师讲授的任一门课程的学生姓名
select st.s_name from student st
where st.s_id not in (
select sc.s_id from score sc
inner join course c on c.c_id=sc.c_id
inner join teacher t on t.t_id=c.t_id and t.t_name=“张三”
)
– 15、查问两门及其以上不及格课程的同学的学号，姓名及其均匀问题
select st.s_id,st.s_name,avg(sc.s_score) from student st
left join score sc on sc.s_id=st.s_id
where sc.s_id in (
select sc.s_id from score sc
where sc.s_score<60 or sc.s_score is NULL
group by sc.s_id having COUNT(sc.s_id)>=2
)
group by st.s_id
– 16、检索 ”01″ 课程分数小于 60，按分数降序排列的学生信息
select st.*,sc.s_score from student st
inner join score sc on sc.s_id=st.s_id and sc.c_id=“01”and sc.s_score<60
order by sc.s_score desc
– 17、按均匀问题从高到低显示所有学生的所有课程的问题以及均匀问题
– 可加 round,case when then else end 使显示更完满
select st.s_id,st.s_name,avg(sc4.s_score)“平均分”,sc.s_score“语文”,sc2.s_score“数学”,sc3.s_score“英语”from student st
left join score sc on sc.s_id=st.s_id and sc.c_id=“01”
left join score sc2 on sc2.s_id=st.s_id and sc2.c_id=“02”
left join score sc3 on sc3.s_id=st.s_id and sc3.c_id=“03”
left join score sc4 on sc4.s_id=st.s_id
group by st.s_id
order by SUM(sc4.s_score) desc
– 18. 查问各科问题最高分、最低分和平均分：以如下模式显示：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，低劣为：80-90，优良为：>=90
select c.c_id,c.c_name,max(sc.s_score)“最高分”,MIN(sc2.s_score)“最低分”,avg(sc3.s_score)“平均分”
,((select count(s_id) from score where s_score>=60 and c_id=c.c_id )/(select count(s_id) from score where c_id=c.c_id))“及格率”
,((select count(s_id) from score where s_score>=70 and s_score<80 and c_id=c.c_id )/(select count(s_id) from score where c_id=c.c_id))“中等率”
,((select count(s_id) from score where s_score>=80 and s_score<90 and c_id=c.c_id )/(select count(s_id) from score where c_id=c.c_id))“优良率”
,((select count(s_id) from score where s_score>=90 and c_id=c.c_id )/(select count(s_id) from score where c_id=c.c_id))“优秀率”
from course c
left join score sc on sc.c_id=c.c_id
left join score sc2 on sc2.c_id=c.c_id
left join score sc3 on sc3.c_id=c.c_id
group by c.c_id
– 19、按各科问题进行排序，并显示排名(实现不齐全)
– mysql 没有 rank 函数
– 加 @score 是为了避免用 union all 后打乱了程序
select c1.s_id,c1.c_id,c1.c_name,@score:=c1.s_score,@i:=@i+1 from (select c.c_name,sc.* from course c
left join score sc on sc.c_id=c.c_id
where c.c_id=“01”order by sc.s_score desc) c1 ,
(select @i:=0) a
union all
select c2.s_id,c2.c_id,c2.c_name,c2.s_score,@ii:=@ii+1 from (select c.c_name,sc.* from course c
left join score sc on sc.c_id=c.c_id
where c.c_id=“02”order by sc.s_score desc) c2 ,
(select @ii:=0) aa
union all
select c3.s_id,c3.c_id,c3.c_name,c3.s_score,@iii:=@iii+1 from (select c.c_name,sc.* from course c
left join score sc on sc.c_id=c.c_id
where c.c_id=“03”order by sc.s_score desc) c3;
set @iii=0;
– 20、查问学生的总成绩并进行排名
select st.s_id,st.s_name
,(case when sum(sc.s_score) is null then 0 else sum(sc.s_score) end)
from student st
left join score sc on sc.s_id=st.s_id
group by st.s_id order by sum(sc.s_score) desc
– 21、查问不同老师所教不同课程平均分从高到低显示
select t.t_id,t.t_name,c.c_name,avg(sc.s_score) from teacher t
left join course c on c.t_id=t.t_id
left join score sc on sc.c_id =c.c_id
group by t.t_id
order by avg(sc.s_score) desc
– 22、查问所有课程的问题第 2 名到第 3 名的学生信息及该课程问题
select a.* from (
select st.,c.c_id,c.c_name,sc.s_score from student st
left join score sc on sc.s_id=st.s_id
inner join course c on c.c_id =sc.c_id and c.c_id=“01”
order by sc.s_score desc LIMIT 1,2 ) a
union all
select b. from (
select st.,c.c_id,c.c_name,sc.s_score from student st
left join score sc on sc.s_id=st.s_id
inner join course c on c.c_id =sc.c_id and c.c_id=“02”
order by sc.s_score desc LIMIT 1,2) b
union all
select c. from (
select st.*,c.c_id,c.c_name,sc.s_score from student st
left join score sc on sc.s_id=st.s_id
inner join course c on c.c_id =sc.c_id and c.c_id=“03”
order by sc.s_score desc LIMIT 1,2) c
– 23、统计各科问题各分数段人数：课程编号, 课程名称,[100-85],[85-70],[70-60],[0-60]及所占百分比
select c.c_id,c.c_name
,((select count(1) from score sc where sc.c_id=c.c_id and sc.s_score<=100 and sc.s_score>80)/(select count(1) from score sc where sc.c_id=c.c_id ))“100-85”
,((select count(1) from score sc where sc.c_id=c.c_id and sc.s_score<=85 and sc.s_score>70)/(select count(1) from score sc where sc.c_id=c.c_id ))“85-70”
,((select count(1) from score sc where sc.c_id=c.c_id and sc.s_score<=70 and sc.s_score>60)/(select count(1) from score sc where sc.c_id=c.c_id ))“70-60”
,((select count(1) from score sc where sc.c_id=c.c_id and sc.s_score<=60 and sc.s_score>=0)/(select count(1) from score sc where sc.c_id=c.c_id ))“60-0”
from course c order by c.c_id
– 24、查问学生均匀问题及其名次
set @i=0;
select a.*,@i:=@i+1 from (
select st.s_id,st.s_name,round((case when avg(sc.s_score) is null then 0 else avg(sc.s_score) end),2)“平均分”from student st
left join score sc on sc.s_id=st.s_id
group by st.s_id order by sc.s_score desc) a
– 25、查问各科问题前三名的记录
select a.* from (
select st.s_id,st.s_name,c.c_id,c.c_name,sc.s_score from student st
left join score sc on sc.s_id=st.s_id
inner join course c on c.c_id=sc.c_id and c.c_id=‘01’
order by sc.s_score desc LIMIT 0,3) a
union all
select b.* from (
select st.s_id,st.s_name,c.c_id,c.c_name,sc.s_score from student st
left join score sc on sc.s_id=st.s_id
inner join course c on c.c_id=sc.c_id and c.c_id=‘02’
order by sc.s_score desc LIMIT 0,3) b
union all
select c.* from (
select st.s_id,st.s_name,c.c_id,c.c_name,sc.s_score from student st
left join score sc on sc.s_id=st.s_id
inner join course c on c.c_id=sc.c_id and c.c_id=‘03’
order by sc.s_score desc LIMIT 0,3) c
– 26、查问每门课程被选修的学生数
select c.c_id,c.c_name,count(1) from course c
left join score sc on sc.c_id=c.c_id
inner join student st on st.s_id=c.c_id
group by st.s_id
– 27、查问出只有两门课程的全副学生的学号和姓名
select st.s_id,st.s_name from student st
left join score sc on sc.s_id=st.s_id
inner join course c on c.c_id=sc.c_id
group by st.s_id having count(1)=2
– 28、查问男生、女生人数
select st.s_sex,count(1) from student st group by st.s_sex
– 29、查问名字中含有 ” 风 ” 字的学生信息
select st.* from student st where st.s_name like“% 风 %”;
– 30、查问同名同性学生名单，并统计同名人数
select st.*,count(1) from student st group by st.s_name,st.s_sex having count(1)>1
– 31、查问 1990 年出世的学生名单
select st.* from student st where st.s_birth like“1990%”;
– 32、查问每门课程的均匀问题，后果按均匀问题降序排列，均匀问题雷同时，按课程编号升序排列
select c.c_id,c.c_name,avg(sc.s_score) from course c
inner join score sc on sc.c_id=c.c_id
group by c.c_id order by avg(sc.s_score) desc,c.c_id asc
– 33、查问均匀问题大于等于 85 的所有学生的学号、姓名和均匀问题
select st.s_id,st.s_name,avg(sc.s_score) from student st
left join score sc on sc.s_id=st.s_id
group by st.s_id having avg(sc.s_score)>=85
– 34、查问课程名称为 ” 数学 ”，且分数低于 60 的学生姓名和分数
select st.s_id,st.s_name,sc.s_score from student st
inner join score sc on sc.s_id=st.s_id and sc.s_score<60
inner join course c on c.c_id=sc.c_id and c.c_name =“数学”
– 35、查问所有学生的课程及分数状况；
select st.s_id,st.s_name,c.c_name,sc.s_score from student st
left join score sc on sc.s_id=st.s_id
left join course c on c.c_id =sc.c_id
order by st.s_id,c.c_name
– 36、查问任何一门课程问题在 70 分以上的姓名、课程名称和分数
select st2.s_id,st2.s_name,c2.c_name,sc2.s_score from student st2
left join score sc2 on sc2.s_id=st2.s_id
left join course c2 on c2.c_id=sc2.c_id
where st2.s_id in(
select st.s_id from student st
left join score sc on sc.s_id=st.s_id
group by st.s_id having min(sc.s_score)>=70)
order by s_id
– 37、查问不及格的课程
select st.s_id,c.c_name,st.s_name,sc.s_score from student st
inner join score sc on sc.s_id=st.s_id and sc.s_score<60
inner join course c on c.c_id=sc.c_id
– 38、查问课程编号为 01 且课程问题在 80 分以上的学生的学号和姓名
select st.s_id,st.s_name,sc.s_score from student st
inner join score sc on sc.s_id=st.s_id and sc.c_id=“01”and sc.s_score>=80
– 39、求每门课程的学生人数
select c.c_id,c.c_name,count(1) from course c
inner join score sc on sc.c_id=c.c_id
group by c.c_id
– 40、查问选修 ” 张三 ” 老师所授课程的学生中，问题最高的学生信息及其问题
select st.*,c.c_name,sc.s_score,t.t_name from student st
inner join score sc on sc.s_id=st.s_id
inner join course c on c.c_id=sc.c_id
inner join teacher t on t.t_id=c.t_id and t.t_name=“张三”
order by sc.s_score desc
limit 0,1
– 41、查问不同课程问题雷同的学生的学生编号、课程编号、学生问题
select st.s_id,st.s_name,sc.c_id,sc.s_score from student st
left join score sc on sc.s_id=st.s_id
left join course c on c.c_id=sc.c_id
where (
select count(1) from student st2
left join score sc2 on sc2.s_id=st2.s_id
left join course c2 on c2.c_id=sc2.c_id
where sc.s_score=sc2.s_score and c.c_id!=c2.c_id
)>1
– 42、查问每门功问题最好的前两名
select a.* from (select st.s_id,st.s_name,c.c_name,sc.s_score from student st
left join score sc on sc.s_id=st.s_id
inner join course c on c.c_id=sc.c_id and c.c_id=“01”
order by sc.s_score desc limit 0,2) a
union all
select b.* from (select st.s_id,st.s_name,c.c_name,sc.s_score from student st
left join score sc on sc.s_id=st.s_id
inner join course c on c.c_id=sc.c_id and c.c_id=“02”
order by sc.s_score desc limit 0,2) b
union all
select c.* from (select st.s_id,st.s_name,c.c_name,sc.s_score from student st
left join score sc on sc.s_id=st.s_id
inner join course c on c.c_id=sc.c_id and c.c_id=“03”
order by sc.s_score desc limit 0,2) c
– 借鉴(更精确, 丑陋):
select a.s_id,a.c_id,a.s_score from score a
where (select COUNT(1) from score b where b.c_id=a.c_id and b.s_score>=a.s_score)<=2 order by a.c_id
– 43、统计每门课程的学生选修人数（超过 5 人的课程才统计）。要求输入课程号和选修人数，查问后果按人数降序排列，
– 若人数雷同，按课程号升序排列
select sc.c_id,count(1) from score sc
left join course c on c.c_id=sc.c_id
group by c.c_id having count(1)>5
order by count(1) desc,sc.c_id asc
– 44、检索至多选修两门课程的学生学号
select st.s_id from student st
left join score sc on sc.s_id=st.s_id
group by st.s_id having count(1)>=2
– 45、查问选修了全副课程的学生信息
select st.* from student st
left join score sc on sc.s_id=st.s_id
group by st.s_id having count(1)=(select count(1) from course)
– 46、查问各学生的年龄
select st.*,timestampdiff(year,st.s_birth,now()) from student st
– 47、查问本周过生日的学生
– 此处可能有问题,week 函数取的为以后年的第几周,2017-12-12 是第 50 周而 2018-12-12 是第 49 周, 能够取月份,day, 星期几(%w),
– 再判断本周是否会继续到下一个月进行判断, 太麻烦, 不会写
select st.* from student st
where week(now())=week(date_format(st.s_birth,‘%Y%m%d’))
– 48、查问下周过生日的学生
select st.* from student st
where week(now())+1=week(date_format(st.s_birth,‘%Y%m%d’))
– 49、查问本月过生日的学生
select st.* from student st
where month(now())=month(date_format(st.s_birth,‘%Y%m%d’))
– 50、查问下月过生日的学生
– 留神: 当 以后月为 12 时, 用 month(now())+ 1 为 13 而不是 1, 可用 timestampadd() 函数或 mod 取模
select st.* from student st
where month(timestampadd(month,1,now()))=month(date_format(st.s_birth,‘%Y%m%d’))
– 或
select st.* from student st where (month(now()) + 1) mod 12 = month(date_format(st.s_birth,‘%Y%m%d’))
