[大数据]50道经典MySQL练习题（含解答）

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数据表介绍

--1.学生表
Student(SId,Sname,Sage,Ssex)
--SId 学?编号,Sname 学?姓名,Sage 出?年?,Ssex 学?性别
--2.课程表
Course(CId,Cname,TId)
--CId 课程编号,Cname 课程名称,TId 教师编号
--3.教师表
Teacher(TId,Tname)
--TId 教师编号,Tname 教师姓名
--4.成绩表
SC(SId,CId,score)
--SId 学?编号,CId 课程编号,score 分数

表中数据如下

-- 学?表 Student
create table Student(SId varchar(10),Sname varchar(10),Sage datetime,Ssex
varchar(10));
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?' , '1990-12-20' , '男');
insert into Student values('04' , '李云' , '1990-12-06' , '男');
insert into Student values('05' , '周梅' , '1991-12-01' , '?');
insert into Student values('06' , '吴兰' , '1992-01-01' , '?');
insert into Student values('07' , '郑?' , '1989-01-01' , '?');
insert into Student values('09' , '张三' , '2017-12-20' , '?');
insert into Student values('10' , '李四' , '2017-12-25' , '?');
insert into Student values('11' , '李四' , '2012-06-06' , '?');
insert into Student values('12' , '赵六' , '2013-06-13' , '?');
insert into Student values('13' , '孙七' , '2014-06-01' , '?');
-- 科?表 Course
create table Course(CId varchar(10),Cname nvarchar(10),TId varchar(10));
insert into Course values('01' , '语?' , '02');
insert into Course values('02' , '数学' , '01');
insert into Course values('03' , '英语' , '03');
-- 教师表 Teacher
create table Teacher(TId varchar(10),Tname varchar(10));
insert into Teacher values('01' , '张三');
insert into Teacher values('02' , '李四');
insert into Teacher values('03' , '王五');
-- 成绩表 SC
create table SC(SId varchar(10),CId varchar(10),score decimal(18,1));
insert into SC values('01' , '01' , 80);
insert into SC values('01' , '02' , 90);
insert into SC values('01' , '03' , 99);
insert into SC values('02' , '01' , 70);
insert into SC values('02' , '02' , 60);
insert into SC values('02' , '03' , 80);
insert into SC values('03' , '01' , 80);
insert into SC values('03' , '02' , 80);
insert into SC values('03' , '03' , 80);
insert into SC values('04' , '01' , 50);
insert into SC values('04' , '02' , 30);
insert into SC values('04' , '03' , 20);
insert into SC values('05' , '01' , 76);
insert into SC values('05' , '02' , 87);
insert into SC values('06' , '01' , 31);
insert into SC values('06' , '03' , 34);
insert into SC values('07' , '02' , 89);
insert into SC values('07' , '03' , 98);

50道练习题目

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 stu.*,s.score FROM
    student AS stu
    JOIN (
            SELECT
                s1.sid,
                s1.score 
            FROM
                ( SELECT sid, score FROM sc WHERE Cid = 01 ) AS s1
                JOIN 
                ( SELECT sid, score FROM sc WHERE Cid = 02 ) AS s2 ON s1.sid = s2.sid 
            WHERE
                s1.score > s2.score 
    ) AS s ON stu.sid = s.sid;

+------+--------+---------------------+------+-------+
| SId  | Sname  | Sage                | Ssex | score |
+------+--------+---------------------+------+-------+
| 02   | 钱电   | 1990-12-21 00:00:00 | 男   |  70.0 |
| 04   | 李云   | 1990-12-06 00:00:00 | 男   |  50.0 |
+------+--------+---------------------+------+-------+

2.查询同时存在" 01 “课程和” 02 "课程的情况

SELECT s1.sid,s1.score as 01_score,s2.score as 02_score FROM
(SELECT sid,score from sc WHERE cid=01) as s1
JOIN
(SELECT sid,score from sc WHERE cid=02) as s2
on s1.sid = s2.sid;

+------+----------+----------+
| sid  | 01_score | 02_score |
+------+----------+----------+
| 01   |     80.0 |     90.0 |
| 02   |     70.0 |     60.0 |
| 03   |     80.0 |     80.0 |
| 04   |     50.0 |     30.0 |
| 05   |     76.0 |     87.0 |
+------+----------+----------+
5 rows in set (0.00 sec)

3.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 )

`mysql SELECT s1.sid,s1.score as 01_score,s2.score as 02_score FROM (SELECT sid,score from sc WHERE cid=01) as s1 LEFT JOIN (SELECT sid,score from sc WHERE cid=02) as s2 on s1.sid = s2.sid;`

4.查询不存在" 01 “课程但存在” 02 "课程的情况

SELECT s2.sid,s1.score as 01_score,s2.score as 02_score FROM
(SELECT sid,score from sc WHERE cid=01) as s1
RIGHT JOIN
(SELECT sid,score from sc WHERE cid=02) as s2
on s1.sid = s2.sid;

5.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 stu.sid,stu.`sname`,s1.avg_score
from student as stu JOIN 
(SELECT sid,avg(score) as avg_score from sc GROUP BY sid) as s1
on stu.sid = s1.sid
WHERE s1.avg_score >60;

6.查询在 SC 表存在成绩的学生信息

SELECT distinct stu.* from student as stu,sc WHERE stu.SId=sc.SId

7.查询所有同学的学?编号、学?姓名、选课总数、所有课程的总成绩(没成绩的显示为 null )

SELECT stu.SId,stu.Sname,count(sc.SId) as '选课总数',sum(sc.score) as '所有课程的总成绩' 
from student as stu LEFT join sc on stu.SId = sc.SId GROUP BY stu.SId,stu.sname;

8.查询「李」姓老师的数量

SELECT count(*) FROM teacher WHERE tname like '李%'

9.查询学过「张三」老师授课的同学的信息

SELECT s1.* FROM 
(SELECT stu.*,sc.CId from student as stu join sc on stu.SId = sc.SId) as s1 
JOIN 
(SELECT teacher.Tname,course.cid FROM course join teacher on course.tid = teacher.Tid) as c1 
on s1.cid = c1.cid WHERE c1.tname = '张三'

10.查询没有学全所有课程的同学的信息

SELECT stu.* FROM student as stu 
where sid not in 
(SELECT s1.sid FROM (SELECT sid,count(sid) as count_sid FROM sc GROUP BY sid) as s1 WHERE s1.count_sid=3)

11.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 stu.* from 
student as stu JOIN sc on stu.sid = sc.SId 
WHERE sc.CId 
in
(SELECT cid FROM sc where sid=01)

12.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 stu.* FROM student as stu JOIN
(SELECT s2.sid FROM sc as s1 JOIN sc as s2 on s1.cid = s2.cid and s1.sid=01 and s2.sid!=01
GROUP BY s2.sid HAVING count(s2.cid)=(SELECT count(*) from sc where sid=01)) as s
on stu.SId = s.sid

13.查询没学过"张三"老师讲授的任一门课程的学生姓名

SELECT * from student  
WHERE SId not in 
(SELECT s1.sid FROM 
(SELECT stu.*,sc.CId from student as stu join sc on stu.SId = sc.SId) as s1 
JOIN 
(SELECT teacher.Tname,course.cid FROM course join teacher on course.tid = teacher.Tid) as c1 
on s1.cid = c1.cid WHERE c1.tname = '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 stu.sname,stu.sid,s1.avg_score from 
student as stu 
JOIN 
(SELECT sid,AVG(score) as avg_score from sc where score<60 GROUP BY sid HAVING count(*)>=2) as s1 
on stu.sid = s1.sid

15.检索" 01 "课程分数小于 60，按分数降序排列的学生信息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 * FROM student WHERE sid in (SELECT sid from sc WHERE cid=01 and score<60 ORDER BY 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 sc.*,s2.avg_score 
FROM sc  
join (SELECT sid,AVG(score) as avg_score from sc GROUP BY sid) as s2
on sc.sid = s2.sid
ORDER BY s2.avg_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 要求输出课程号和选修?数，查询结果按?数降序排列，若人数相同，按课程号升序排列

SELECT sc.cid,course.Cname,max(sc.score) as '最高分',min(sc.score) as '最低分',
AVG(sc.score) as '平均分',count(sc.CId) as '选修人数',
SUM(case when sc.score>=60 then 1 else 0 end)/count(sc.CId) as '及格率',
SUM(case when sc.score>=70 and sc.score<80 then 1 else 0 end)/count(sc.CId) as '中等率',
SUM(case when sc.score>=80 and sc.score<90 then 1 else 0 end)/count(sc.CId) as '优良率',
SUM(case when sc.score>=90 then 1 else 0 end)/count(sc.CId) as '优秀率'
from sc,course WHERE sc.CId=course.CId 
GROUP BY cid,course.Cname
ORDER BY '选修人数' DESC,sc.cid;

18.按各科平均成绩进行排序，并显示排名， Score 重复时保留名次空缺

select s2.cid,s2.avg_sc,count(s1.avg_sc) as rank
from 
(SELECT cid,ROUND(AVG(score),2) as avg_sc from sc GROUP BY cid ) as s1
join
(SELECT cid,ROUND(AVG(score),2) as avg_sc from sc GROUP BY cid ) as s2
on s1.avg_sc>=s2.avg_sc and s1.cid = s1.cid
group by s2.cid, s2.avg_sc
order by rank;

19.按各科平均成绩进行排序，并显示排名， Score 重复时不保留名次空缺

SELECT b.cid,b.avg_sc,@i:=@i+1 as rank 
from (SELECT @i :=0) as a,
(SELECT cid,round(avg(score),2) as avg_sc from sc GROUP BY cid ORDER BY avg_sc desc) as b

20.查询学生的总成绩，并进行排名，总分重复时保留名次空缺

SELECT s2.sid,s2.sum_sc,COUNT(s2.sum_sc) as rank from
(SELECT sid,sum(score) as sum_sc from sc GROUP BY sid ORDER BY sid) as s1
JOIN 
(SELECT sid,sum(score) as sum_sc from sc GROUP BY sid ORDER BY sid) as s2
on s1.sum_sc>=s2.sum_sc
group by s2.sid,s2.sum_sc
order by rank;

21.查询学生的总成绩，并进行排名，总分重复时不保留名次空缺

SELECT b.sid,b.sum_sc,@i:=@i+1 as rank 
from (SELECT @i :=0) as a,
(SELECT sid,sum(score) as sum_sc from sc GROUP BY sid ORDER BY sum_sc desc) as b

22.统计各科成绩各分数段人数：课程编号，课程名称，[100-85]，[85-70]，[70-60]，[60-0]及所占百分比

SELECT sc.CId,c.cname,
SUM(case when sc.score>85 and sc.score<=100 then 1 else 0 end) as '[100-85]',
SUM(case when sc.score>85 and sc.score<=100 then 1 else 0 end)/count(sc.CId) as '百分比',
SUM(case when sc.score>70 and sc.score<=85 then 1 else 0 end) as '[85-70]',
SUM(case when sc.score>70 and sc.score<=85 then 1 else 0 end)/count(sc.CId) as '百分比',
SUM(case when sc.score>60 and sc.score<=70 then 1 else 0 end) as '[70-60]',
SUM(case when sc.score>60 and sc.score<=70 then 1 else 0 end)/count(sc.CId) as '百分比',
SUM(case when sc.score>0 and sc.score<=60 then 1 else 0 end) as '[60-0]',
SUM(case when sc.score>0 and sc.score<=60 then 1 else 0 end)/count(sc.CId) as '百分比'
FROM sc join course as c on sc.CId=c.cid
GROUP BY sc.CId,c.cname

23.查询各科成绩前三名的记录

(select CId,score from SC where CId = '01' order by score desc limit 3)
union all
(select CId,score from SC where CId = '02' order by score desc limit 3)
union all
(select CId,score from SC where CId = '03' order by score desc limit 3)

24.查询每门课程被选修的学生数

SELECT cid,count(cid) as 选课人数 FROM sc GROUP BY CId;

25.查询出只选修两门课程的学生学号和姓名

SELECT s2.sid,s2.sname FROM
(SELECT sid,count(sid) as '选修课程数'  FROM sc GROUP BY SId) as s1
JOIN student as s2 on s1.SId = s2.SId
WHERE s1.选修课程数=2

26.查询男生、女生人数

SELECT ssex,count(SId) FROM student GROUP BY ssex

27.查询名字中含有「风」字的学生信息

SELECT * FROM student WHERE Sname like '%风%'

28.查询同名同姓学生名单，并统计同名?数

select sname,count(*) as 人数
from student
group by sname
having count(*)>=2;

29.查询 1990 年出生的学生名单

SELECT * FROM student WHERE YEAR(sage)=1990

30.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 cid,AVG(score) FROM sc GROUP BY CId ORDER BY AVG(score),CId

31.查询平均成绩?于等于 85 的所有学生的学号、姓名和平均成绩

SELECT s1.sid,s1.sname,s2.平均分
FROM student as s1 
join
(SELECT sid,AVG(score) as '平均分' FROM sc GROUP BY sid) as s2
on s1.SId = s2.SId
WHERE s2.平均分>=85

32.查询课程名称为「数学」，且分数低于 60 的学生姓名和分数

SELECT s2.sname,s1.score
FROM
course as c1 JOIN sc as s1 on c1.cid=s1.CId
JOIN student as s2 on s1.sid =s2.SId
WHERE c1.cname='数学' and s1.score<60

33.查询所有学生的课程及分数情况（存在学生没成绩，没选课的情况）

SELECT s2.sname,s2.sid,c1.cname,s1.score
FROM
 sc as s1
right JOIN student as s2 on s1.sid =s2.SId
left JOIN course as c1 on s1.cid=c1.cid
GROUP BY s2.sid,s2.sname,c1.cname,s1.score

34.查询任何?门课程成绩在 70 分以上的姓名、课程名称和分数

SELECT s2.sname,c1.cname,s1.score
FROM
 sc as s1
JOIN student as s2 on s1.sid =s2.SId
JOIN course as c1 on s1.cid=c1.cid
WHERE s1.score>70

35.查询不及格的课程

SELECT s2.sname,c1.cname,s1.score
FROM
 sc as s1
JOIN student as s2 on s1.sid =s2.SId
JOIN course as c1 on s1.cid=c1.cid
WHERE s1.score<60

36.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 s1.sid,s2.Sname
FROM
 (SELECT sid,score FROM sc WHERE cid='01') as s1
JOIN student as s2 on s1.sid =s2.SId
WHERE score>=80


select SC.SId,Student.Sname
from SC join Student on SC.SId = Student.SId
where SC.Score > 80 and SC.CId = '01';

37.求每门课程的学生人数

SELECT cid,count(cid) as '学生人数' FROM sc GROUP BY cid

38.成绩不重复，查询选修「张三」老师所授课程的学生中，成绩最高的学生信息及其成绩

SELECT s2.sname,s2.SId,s1.score
FROM
course as c1 JOIN sc as s1 on c1.cid=s1.CId
JOIN student as s2 on s1.sid =s2.SId
JOIN teacher as t1 on c1.tid=t1.tid
WHERE t1.tname='张三'
ORDER BY s1.score DESC  LIMIT 1

39.成绩有重复的情况下，查询选修「张三」老师所授课程的学生中，成绩最高的学生信息及其成绩

SELECT s2.sname,s2.sid,s1.score FROM student as s2 JOIN sc as s1 on s1.SId =s2.SId
WHERE score=(
SELECT max(score) FROM sc as s1 WHERE cid=(SELECT c1.cid FROM course as c1 JOIN teacher as t1 on c1.tid = t1.tid WHERE t1.tname = '张三'))

-- 其它?案
select a.sname,b.score
from student a join sc b on a.sid=b.sid
and b.cid in (select cid from course where tid in (select tid from teacher
where tname='张三'))
join (select cid,max(score) m from sc group by cid) c on b.cid=c.cid and
b.score=c.m;

40.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 sid,cid,score FROM sc WHERE sid=(select sid from (select sid,score from sc group by sid,score) as s1
group by sid having count(sid)=1)

-- 其他方案
select distinct s1.SId,s1.CId,s1.Score
from SC s1 join SC s2
on s1.CId != s2.CId and s1.score = s2.score
group by s1.SId,s1.CId,s1.Score

41.查询每门课程成绩最好的前两名

SELECT s1.* FROM sc s1 WHERE
(
SELECT COUNT(1) FROM sc s2 WHERE
s1.cid=s2.cid AND s2.score>=s1.score
)<=2
ORDER BY s1.cid,s1.score DESC;

-- 其他方案
(select CId,score from SC where CId = '01' order by score desc limit 2)
union all
(select CId,score from SC where CId = '02' order by score desc limit 2)
union all
(select CId,score from SC where CId = '03' order by score desc limit 2)

42.统计每门课程的学生选修人数（超过 5 ?的课程才统计）。

SELECT cid,count(cid) as '学生人数' FROM sc GROUP BY cid HAVING count(cid)>5

43.检索至少选修两门课程的学生学号

SELECT cid,count(cid) as '学生人数' FROM sc GROUP BY cid HAVING count(cid)>=5

44.查询选修了全部课程的学生信息

SELECT sid FROM sc GROUP BY SId HAVING count(sid)=3

45.查询各学生的年龄，只按年份来算

SELECT sname,year(now())-YEAR(sage) as '年龄' FROM student

46.按照出生日期来算，当前月日 < 出生年月的月日则，年龄减?

TIMESTAMPDIFF() 从日期时间表达式中减去间隔 https://dev.mysql.com/doc/refman/5.7/en/date-and-time-functions.html

select student.sid, student.sname,student.ssex, sage,
timestampdiff(year,sage,now()) as '按月日计算',  
year(now())-year(sage) as '按年份计算'  
from student;

47.查询本周过生日的学生

返回日期从范围内的数字日历星期1到53

select sid, sname,ssex, sage
from student
WHERE WEEKOFYEAR(student.Sage)=WEEKOFYEAR(CURDATE())

48.查询下周过生日的学生

select sid, sname,ssex, sage
from student
where WEEKOFYEAR(student.Sage)=WEEKOFYEAR(CURDATE())+1

49.查询本月过生日的学生

select *
from student
where month(student.sage)=month(now());

50.查询下月过生日的学生

select *
from student
where month(student.sage)=month(now())+1;

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