> 50道SQL练习题 - Yuyy
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50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 denied for 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 )

SELECT Student.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)= (select count(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
    http://yuyy.info/image/深度截图_选择区域_20200330113242.png
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<80 then 1 else 0 end)/count(1) as 中等率,
    sum(case when sc.score>=80 and sc.score<90 then 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) as rank
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) as total
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>=60 then 1 else 0 end)as '60-70',
    sum(case when score<70 and score>=60 then 1 else 0 end)/count(1)*100 as '60-70百分比',
    sum(case when score<85 and score>=70 then 1 else 0 end)as '70-85',
    sum(case when score<85 and score>=70 then 1 else 0 end)/count(1)*100 as '70-85百分比',
    sum(case when score<=100 and score>=85 then 1 else 0 end)as '85-100',
    sum(case when score<=100 and score>=85 then 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 and a.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
from teacher,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

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;

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50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 denied for user 'root'@'%' to database 'xxx' 解决方法:grant all on xxx.* to 'root'@'%' identified by …
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2020-03-25
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