​​45道SQL练习题！

 

数据库准备

·        在服务器上创建好数据库后需要配置访问权限，否则会报错Access deniedfor user 'root'@'%' to database 'xxx'

·        解决方法：grant all on xxx.* to 'root'@'%'identified by 'password' with grant option;

1.学生表

Student(SId,Sname,Sage,Ssex)
–SId 学生编号,Sname 学生姓名,Sage 出生年月,Ssex 学生性别

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' , '孙风' , '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', '女');

insert into Student values('06' , '吴兰' , '1992-01-01', '女');

insert into Student values('07' , '郑竹' , '1989-01-01', '女');

insert into Student values('09' , '张三' , '2017-12-20', '女');

insert into Student values('10' , '李四' , '2017-12-25', '女');

insert into Student values('11' , '李四' , '2012-06-06', '女');

insert into Student values('12' , '赵六' , '2013-06-13', '女');

insert into Student values('13' , '孙七' , '2014-06-01', '女');

2.课程表

Course(CId,Cname,TId)
–CId 课程编号,Cname 课程名称,TId 教师编号

create table Course(CId varchar(10),Cname nvarchar(10),TId varchar(10));

insert into Course values('01' , '语文' , '02');

insert into Course values('02' , '数学' , '01');

insert into Course values('03' , '英语' , '03');

3.教师表

Teacher(TId,Tname)
–TId 教师编号,Tname 教师姓名

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');

4.成绩表

SC(SId,CId,score)
–SId 学生编号,CId 课程编号,score 分数

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);

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 Student.*,01Score,02Score

from Student,

    (select tb1.SId,01Score,02Score

    from

        (select SId,score as 01Score

        from sc

        where CId='01')as tb1,

        (select SId,score as 02Score

        from sc

        where CId='02')as tb2

    where tb1.SId=tb2.SId and 01Score>02Score)as tb3

where tb3.SId=Student.SId

1.1 查询同时存在” 01 “课程和” 02 “课程的情况

SELECT Student.*

FROM Student JOIN

    (SELECT tb1.SId

    FROM

        (SELECT SId

        FROM sc

        WHERE CId='01') as tb1 join

        (SELECT SId

        FROM sc

        WHERE CId='01') as tb2 on tb1.SId=tb2.SId) AS tb3

    ON tb3.SId=Student.SId

1.2 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 )

可能不存在？？？“可能”可以作为条件吗？

1.3 查询不存在” 01 “课程但存在” 02 “课程的情况

SELECT  *

FROM sc

where   SId not in(SELECT SId

    from sc

    where CId='01')

    and CId='02'

2.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 Student.Sname,tb1.*

from Student join

    (select avg(score) as avg_score,SId

    from sc

    group by SId

    HAVING avg(score)>60)as tb1

    on Student.SId=tb1.SId

3.查询在 SC 表存在成绩的学生信息

select distinct Student.*

from Student JOIN

    sc on Student.SId=sc.SId

4.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(没成绩的显示为 null )

SELECTStudent.Sname,Student.SId,ifnull(tb1.class_num,0),tb1.class_score

from Student left join

    (select SId,count(1) as class_num,sum(score)as class_score

    from sc

    group by SId)as tb1

    on Student.SId=tb1.SId

4.1查有成绩的学生信息

select Student.*

from Student

where Student.SId in

    (select SId

    from sc)

5.查询「李」姓老师的数量

select count(1)

from Teacher

where Tname like'李%'

6.查询学过「张三」老师授课的同学的信息

·        多表联合查询时，如果条件简单（没有分组之内的），可以考虑不用子查询

·        错误例子

select Student.*

from Student join

    (select SId

    from sc

    where CId=

        (select CId

        from Course

        where TId=

           (select TId

           from Teacher

           where Tname like '张三')))as tb1

    on Student.SId=tb1.SId

·        正确例子

select student.* from student,teacher,course,sc

where

   student.sid= sc.sid

    and course.cid=sc.cid

    and course.tid = teacher.tid

    and tname = '张三';

7.查询没有学全所有课程的同学的信息

·        没仔细思考需要查出的记录由哪几部分组成，没有选全包括选了部分，这个可以通过sc表得出，还包括没选课的，这个就需要计算了。

·        错误例子

select sid

from sc

group by sid

HAVING count(1)<

    (select count(1)as class_num

    from Course)

·        正确例子

select * from student

where student.sid not in (

  select sc.sid from sc

  group by sc.sid

  having count(sc.cid)= (selectcount(cid) from course)

);

8.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

·        祥哥说的：少用子查询，效率低。

·        错误例子

select * from student

where student.sid in (

    select sc.sid from sc

    where sc.cid in(

        select sc.cid from sc

        where sc.sid = '01'

    )

);

·        正确例子

select distinct student.*

from sc,student

where cid in(

    select cid

    from sc

    where sid='01')

    and sc.sid=student.sid

9.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 sid

from sc

where cid in(

    SELECT cid

    from sc

    where sid='01')

    and sid <> '01'

group by sid

having count(1)=(

    select count(1)

    from sc

    where sid='01')

10.查询没学过”张三”老师讲授的任一门课程的学生姓名

·        每张表尽量只查询一次

select Student.*

from Student

where SId not in(

        select sc.sid

        from sc,teacher,course

        where tname = '张三'

           and teacher.tid=course.tid

           and course.cid=sc.cid)

11.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 student.sid,avg(score),sname

from sc,student

where score<60

    and student.SId=sc.sid

group by sid

having count(1)>1

12.检索” 01 “课程分数小于 60，按分数降序排列的学生信息

select student.*,score

from sc,student

where cid='01'

    and student.sid=sc.sid

    and score<60

order by score desc

13.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

·        这里同时使用了多个join

·        效果还不错

​

select student.sid,student.sname,tb1.cname,tb1.score,tb2.cname,tb2.score,tb3.cname,tb3.score,tb4.avg_score

from student

    left join

    (select sc.sid,cname,score

    from sc,course

    where sc.cid=course.cid

        and course.cname='语文')as tb1

    on student.sid=tb1.sid

    left join

    (select sc.sid,cname,score

    from sc,course

    where sc.cid=course.cid

        and course.cname='数学')as tb2

    on student.sid=tb2.sid

    left join

    (select sc.sid,cname,score

    from sc,course

    where sc.cid=course.cid

        and course.cname='英语')as tb3

    on student.sid=tb3.sid

    left JOIN

    (select sid,avg(score) as avg_score

    from sc

    group by sid) as tb4

    on student.sid=tb4.sid

order by tb4.avg_score desc

14.查询各科成绩最高分、最低分和平均分：

select cname,max(score),min(score),avg(score)

from sc,course

where sc.cid=course.cid

group by sc.cid

14.1.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

·        利用sum，case when求及格率，中等率，优良率，优秀率

select cname,max(score),min(score),avg(score),

    sum(case when sc.score>=60 then 1 else 0 end)/count(1) as 及格率,

    sum(case when sc.score>=70 and sc.score<80then 1 else 0 end)/count(1) as 中等率,

    sum(case when sc.score>=80 and sc.score<90then 1 else 0 end)/count(1) as 优良率,

    sum(case when sc.score>=90 then 1 else 0 end)/count(1) as 优秀率

from sc,course

where sc.cid=course.cid

group by sc.cid

14.2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

·        多条件排序

select cid,count(1)

from sc

group by cid

order by COUNT(1) desc,cid

15.按各科成绩进行排序，并显示排名， Score 重复时保留名次空缺

·        利用自交，达到排名的效果

select tb1.cid,tb1.sid,tb1.score,count(tb2.cid) asrank

from sc as tb1 left join sc as tb2 on tb1.score<tb2.score and tb1.cid=tb2.cid

group by tb1.cid,tb1.sid

order by cid,rank asc;

15.1 按各科成绩进行排序，并显示排名， Score 重复时合并名次

16.查询学生的总成绩，并进行排名，总分重复时保留名次空缺

·        使用变量以及if

set @r=0;

set @old=-1;

select Sname,tb1.total,tb1.rank

from(

select tb.*,if(@old<>tb.total,@r:=@r+1,@r)as rank,@old:=tb.total

from (

select sc.sid,sum(sc.score) astotal

from sc

group by sc.sid

order by total desc)as tb)as tb1,student

where student.SId=tb1.sid

ORDER BY tb1.rank

16.1 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

17.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

·        利用sum,case when求各分数段的人数以及百分比

select cname,tb.*

from course,(

select sc.cid,

    sum(case when score<60 then 1 else 0 end)as '0-60',

    sum(case when score<60 then 1 else 0 end)/count(1)*100 as '0-60百分比',

    sum(case when score<70 and score>=60then 1 else 0 end)as '60-70',

    sum(case when score<70 and score>=60then 1 else 0 end)/count(1)*100 as '60-70百分比',

    sum(case when score<85 and score>=70then 1 else 0 end)as '70-85',

    sum(case when score<85 and score>=70then 1 else 0 end)/count(1)*100 as '70-85百分比',

    sum(case when score<=100 and score>=85then 1 else 0 end)as '85-100',

    sum(case when score<=100 and score>=85then 1 else 0 end)/count(1)*100 as '85-100百分比'

from sc

group by sc.cid)as tb

where tb.cid=course.cid

18.查询各科成绩前三名的记录

·        利用where,count找前三名

select sc.*

from sc

where (

select count(1)

from sc as tb

where sc.cid=tb.cid and sc.score<tb.score

)<3

order by sc.cid,sc.score desc

·        利用自交找前三名

select a.sid,a.cid,a.score from sc a

left join sc b on a.cid = b.cid anda.score<b.score

group by a.cid, a.sid

having count(b.cid)<3

order by a.cid;

19.查询每门课程被选修的学生数

select cid, count(sid)

from sc

group by cid;

20.查询出只选修两门课程的学生学号和姓名

select sname, count(cid)

from sc ,student

where sc.sid=student.sid

group by sc.sid

having count(cid)=2;

21.查询男生、女生人数

·        count(表达式)

select count(Ssex='男')as '男',count(Ssex='女') as '女'

from student

22.查询名字中含有「风」字的学生信息

select *

from student

where sname like '%风%'

23.查询同名同性学生名单，并统计同名人数

·        同名同姓：group by

select student.*,tb.count

from student,(

select sname,count(1)as count

from student

group by sname

having count(1)>1)as tb

where student.sname=tb.sname

24.查询 1990 年出生的学生名单

·        日期比较，可以使用字符串

select *

from student

where Sage between '1990-0-0 00:00:00' and '1991-0-0 00:00:00'

25.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) desc,cid

26.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 sname,student.sid,avg(score)

from sc,student

where sc.sid=student.sid

group by sc.sid

27.查询课程名称为「数学」，且分数低于 60 的学生姓名和分数

select sname,score

from student,sc,course

where cname='数学'

and student.sid=sc.sid

and course.cid=sc.cid

and score<60

28.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）

·        当遇到“所有”这样的字眼时，应该考虑使用join

select sname,cname,score

from student left join sc on student.sid=sc.sid left join course on course.cid=sc.cid

29.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

select sname,cname,score

from course left join sc on course.cid=sc.cid left join student  on student.sid=sc.sid

where score>70

30.查询不及格的课程

select sname,cname,score

from course left join sc on course.cid=sc.cid left join student  on student.sid=sc.sid

where score<60

31.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 sc.sid,sname,score

from sc,student

where sc.sid=student.sid

and sc.cid = '01'

and score >80

32.求每门课程的学生人数

select cname,count(sid)

from sc,course

where sc.cid=course.cid

group by sc.cid

33.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

·        最高：limit

select sname,score

fromteacher,sc,student,course

where tname='张三'

and teacher.tid=course.tid

and course.cid=sc.cid

and student.sid=sc.sid

order by score desc

limit 1

34.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

35.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 sc.*

from sc,(

select score,sid

from sc

group by score,sid

having count(score)>1 ) as tb

where sc.score=tb.score and sc.sid=tb.sid

36.查询每门功成绩最好的前两名

·        同18题

37.统计每门课程的学生选修人数（超过 5 人的课程才统计）。

select sc.cid,count(1)

from course left join sc on course.cid=sc.cid

group by sc.cid

having count(1)>5

38.检索至少选修两门课程的学生学号

select sid

from sc

group by sid

having count(1)>=2

39.查询选修了全部课程的学生信息

select sid

from sc

group by sid

having count(1)=(select count(1) from course)

40.查询各学生的年龄，只按年份来算

41.按照出生日期来算，当前月日 < 出生年月的月日则，年龄减一

·        日期计算应使用函数TIMESTAMPDIFF

select *,TIMESTAMPDIFF(year,Sage,current_date)as age

from student

42.查询本周过生日的学生

·        WEEKOFYEAR

select *

from student

where WEEKOFYEAR(student.Sage)=WEEKOFYEAR(CURDATE());

43.查询下周过生日的学生

select *

from student

where WEEKOFYEAR(student.Sage)=WEEKOFYEAR(CURDATE())+1;

44.查询本月过生日的学生

·        MONTH

select *

from student

where MONTH(student.Sage)=MONTH(CURDATE());

45.查询下月过生日的学生

select *

from student

where MONTH(student.Sage)=MONTH(CURDATE())+1;

如果大家有其他的方法，或者想要知道哪方面的办公技巧，下方评论哦~

学习资料见知识星球。

以上就是今天要分享的技巧，你学会了吗？若有什么问题，欢迎在下方留言。

快来试试吧，小琥 my21ke007。获取 1000个免费 Excel模板福利​​​​！

更多技巧， www.excelbook.cn

欢迎 加入 零售创新 知识星球，知识星球主要以数据分析、报告分享、数据工具讨论为主；

1、价值上万元的专业的PPT报告模板。

2、专业案例分析和解读笔记。

3、实用的Excel、Word、PPT技巧。

4、VIP讨论群，共享资源。

5、优惠的会员商品。

6、一次付费只需99元，即可下载本站文章涉及的文件和软件。