SQL经典50题及答案解析

Excel处理数据的级别是万条以内,当超过10万条以上时在exce里操作是极其困难的,这时候就需要我们使用SQL来查询,俗称取数。对于SQl的学习建议在了解基本逻辑后立刻进入项目练习,以下是来源于网络的sql练习题目及答案解析

1.SQL执行顺序
SQL执行顺序

2、建表语句

create table Student(sid varchar(10),sname varchar(10),sage datetime,ssex nvarchar(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' , '1990-05-20' , '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');
create table Course(cid varchar(10),cname varchar(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');
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');
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);

3、表结构预览

–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
–课程表
Course(CId,Cname,TId)
–CId 课程编号,Cname 课程名称,TId 教师编号
–教师表
Teacher(TId,Tname)
–TId 教师编号,Tname 教师姓名
–成绩表
SC(SId,CId,score)
–SId 学生编号,CId 课程编号,score 分数

题目

1、查询“01”课程比“02”课程成绩高的所有学生的学号;
2、查询平均成绩大于60分的同学的学号和平均成绩;
3、查询所有同学的学号、姓名、选课数、总成绩
4、查询姓“李”的老师的个数;
5、查询没学过“张三”老师课的同学的学号、姓名;
6、查询学过编号“01”并且也学过编号“02”课程的同学的学号、姓名;
7、查询学过“张三”老师所教的课的同学的学号、姓名;
8、查询课程编号“01”的成绩比课程编号“02”课程低的所有同学的学号、姓名;
9、查询所有课程成绩小于60分的同学的学号、姓名;
10、查询没有学全所有课的同学的学号、姓名;
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学号和姓名;
12、查询和”01″号的同学学习的课程完全相同的其他同学的学号和姓名
13、把“SC”表中“张三”老师教的课的成绩都更改为此课程的平均成绩;
14、查询没学过”张三”老师讲授的任一门课程的学生姓名
15、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
16、检索”01″课程分数小于60,按分数降序排列的学生信息
17、按平均成绩从高到低显示所有学生的平均成绩
18、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
19、按各科平均成绩从低到高和及格率的百分数从高到低顺序
20、查询学生的总成绩并进行排名
21、查询不同老师所教不同课程平均分从高到低显示
22、查询所有课程的成绩第2名到第3名的学生信息及该课程成绩
23、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],[0-60]及所占百分比
24、查询学生平均成绩及其名次
25、查询各科成绩前三名的记录
26、查询每门课程被选修的学生数
27、查询出只选修了一门课程的全部学生的学号和姓名
28、查询男生、女生人数
29、查询名字中含有”风”字的学生信息
30、查询同名同性学生名单,并统计同名人数
31、查询1990年出生的学生名单(注:Student表中Sage列的类型是datetime)
32、查询每门课程的平均成绩,结果按平均成绩升序排列,平均成绩相同时,按课程号降序排列
37、查询不及格的课程,并按课程号从大到小排列
38、查询课程编号为”01″且课程成绩在60分以上的学生的学号和姓名;
40、查询选修“张三”老师所授课程的学生中,成绩最高的学生姓名及其成绩
42、查询每门功课成绩最好的前两名
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
44、检索至少选修两门课程的学生学号
45、查询选修了全部课程的学生信息
46、查询各学生的年龄
47、查询本周过生日的学生
48、查询下周过生日的学生
49、查询本月过生日的学生
50、查询下月过生日的学生






参考答案

1、查询“01”课程比“02”课程成绩高的所有学生的学号;

select distinct t1.sid as sid
from 
    (select * from sc where cid='01')t1
left join 
    (select * from sc where cid='02')t2
on t1.sid=t2.sid
where t1.score>t2.score
2、查询平均成绩大于60分的同学的学号和平均成绩;

select 
    sid
    ,avg(score)
from sc
group by sid
having avg(score)>60
3、查询所有同学的学号、姓名、选课数、总成绩

select
    student.sid as sid
    ,sname
    ,count(distinct cid) course_cnt
    ,sum(score) as total_score
from student
left join sc 
on student.sid=sc.sid
group by sid,sname
4、查询姓“李”的老师的个数;

select
    count(distinct tid) as teacher_cnt
from teacher
where tname like '李%'
5、查询没学过“张三”老师课的同学的学号、姓名;

select
    sid,sname
from student
where sid not in 
    (
        select
            sc.sid
        from teacher
        left join course
            on teacher.tid=course.tid
        left join sc
            on course.cid=sc.cid
        where teacher.tname='张三'
    )
6、查询学过“01”并且也学过编号“02”课程的同学的学号、姓名;

select
    t.sid as sid
    ,sname
from 
    (
        select
            sid
            ,count(if(cid='01',score,null)) as count1
            ,count(if(cid='02',score,null)) as count2
        from sc
        group by sid
        having count(if(cid='01',score,null))>0 and count(if(cid='02',score,null))>0
    )t
left join student
    on t.sid=student.sid
7、查询学过“张三”老师所教的课的同学的学号、姓名;

select
    student.sid
    ,sname
from 
    (
        select
            distinct cid 
        from course
        left join teacher 
        on course.tid=teacher.tid
        where teacher.tname='张三'
    )course
left join sc 
    on course.cid=sc.cid
left join student
    on sc.sid=student.sid
group by student.sid,sname
8、查询课程编号“01”的成绩比课程编号“02”课程低的所有同学的学号、姓名;

select
    t1.sid,sname
from 
    (
        select distinct t1.sid as sid
        from 
            (select * from sc where cid='01')t1
        left join 
            (select * from sc where cid='02')t2
        on t1.sid=t2.sid
        where t1.score>t2.score
    )t1
left join student
    on t1.sid=student.sid
9、查询所有课程成绩小于60分的同学的学号、姓名;

select
    t1.sid,sname
from 
    (
        select
            sid,max(score)
        from sc
        group by sid
        having max(score<60)
    )t1
left join student
    on t1.sid=student.sid
10、查询没有学全所有课的同学的学号、姓名;

select
    t1.sid,sname
from 
    (
        select
            count(cid),sid
        from sc
        group by sid
        having count(cid) < (select count(distinct cid) from course)
    )t1
left join student
    on t1.sid=student.sid
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学号和姓名;

select
    distinct sc.sid
from 
    (
        select
            cid
        from sc
        where sid='01'
    )t1
left join sc
    on t1.cid=sc.cid
12、查询和”01″号的同学学习的课程完全相同的其他同学的学号和姓名

#注意是和'01'号同学课程完全相同但非学习课程数相同的,这里我用左连接解决这个问题
select
    t1.sid,sname
from
    (
        select
            sc.sid
            ,count(distinct sc.cid)
        from 
            (
                select
                    cid
                from sc
                where sid='01'
            )t1 #选出01的同学所学的课程
        left join sc
            on t1.cid=sc.cid
        group by sc.sid
        having count(distinct sc.cid)= (select count(distinct cid) from sc where sid = '01')
    )t1
left join student
    on t1.sid=student.sid
where t1.sid!='01'
13、把“SC”表中“张三”老师教的课的成绩都更改为此课程的平均成绩;

#暂跳过update题目
14、查询没学过”张三”老师讲授的任一门课程的学生姓名

select 
    sname
from student
where sid not in
    (
        select
            distinct sid
        from sc
        left join course
            on sc.cid=course.cid
        left join teacher
            on course.tid=teacher.tid 
        where tname='张三'
    )
15、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
    t1.sid,sname,avg_score
from 
    (
        select
            sid,count(if(score<60,cid,null)),avg(score) as avg_score
        from sc
        group by sid
        having count(if(score<60,cid,null)) >=2
    )t1
left join student
    on t1.sid=student.sid
16、检索”01″课程分数小于60,按分数降序排列的学生信息

select 
    sid,if(cid='01',score,100)
from sc
where if(cid='01',score,100)<60
order by if(cid='01',score,100) desc
17、按平均成绩从高到低显示所有学生的平均成绩

select sid,avg(score)
from sc
group by sid
order by avg(score) desc
18、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

select
    sc.cid
    ,cname
    ,max(score) as max_score
    ,min(score) as min_score
    ,avg(score) as avg_score
    ,count(if(score>=60,sid,null))/count(sid) as pass_rate
from sc 
left join course
    on sc.cid=course.cid
group by sc.cid
19、按各科平均成绩从低到高和及格率的百分数从高到低顺序

#这里先按照平均成绩排序,再按照及格百分数排序,题目有点奇怪
select 
    cid
    ,avg(score) as avg_score
    ,count(if(score>=60,sid,null))/count(sid) as pass_rate
from sc
group by cid
order by avg_score,pass_rate desc
20、查询学生的总成绩并进行排名

select
    sid
    ,sum(score) as sum_score
from sc
group by sid
order by sum_score desc
21、查询不同老师所教不同课程平均分从高到低显示

select
    tid
    ,avg(score) as avg_score
from course
left join sc
    on course.cid=sc.cid
group by tid
order by avg_score desc
22、查询所有课程的成绩第2名到第3名的学生信息及该课程成绩

select
    sid,rank_num,score,cid
from
    (
        select
            rank() over(partition by cid order by score desc) as rank_num
            ,sid
            ,score
            ,cid
        from sc
    )t
where rank_num in (2,3)
23、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],[0-60]及所占百分比

select
    sc.cid
    ,cname
    ,count(if(score between 85 and 100,sid,null))/count(sid)
    ,count(if(score between 70 and 85,sid,null))/count(sid)
    ,count(if(score between 60 and 70,sid,null))/count(sid)
    ,count(if(score between 0 and 60,sid,null))/count(sid)
from sc
left join course
    on sc.cid=course.cid
group by sc.cid,cname
24、查询学生平均成绩及其名次

select
    sid
    ,avg_score
    ,rank() over (order by avg_score desc)
from 
    (
        select
            sid
            ,avg(score) as avg_score
        from sc
        group by sid
    )t
25、查询各科成绩前三名的记录

select
    sid,cid,rank1
from 
    (
        select
            cid
            ,sid
            ,rank() over(partition by cid order by score desc) as rank1
        from sc
    )t
where rank1<=3
26、查询每门课程被选修的学生数

select
    count(sid)
    ,cid
from sc
group by cid
27、查询出只选修了一门课程的全部学生的学号和姓名

#只查出来sid即可,后面懒得交student表
select
    sid
from sc
group by sid
having count(cid) =1
28、查询男生、女生人数

select
    ssex
    ,count(distinct sid)
from student
group by ssex
29、查询名字中含有”风”字的学生信息

select
    sid,sname
from student
where sname like '%风%'
30、查询同名同性学生名单,并统计同名人数

#题目有歧义,这套题的质量感觉有点差
select
    ssex
    ,sname
    ,count(sid)
from student
group by ssex,sname
having count(sid)>=2
31、查询1990年出生的学生名单(注:Student表中Sage列的类型是datetime)

select
    sid,sname,sage
from student
where year(sage)=1990
32、查询每门课程的平均成绩,结果按平均成绩升序排列,平均成绩相同时,按课程号降序排列

select
    cid,avg(score) as avg_score
from sc
group by cid
order by avg_score,cid desc
37、查询不及格的课程,并按课程号从大到小排列

#有问题的题目
select
    cid,sid,score
from sc
where score<60
order by cid desc,sid
38、查询课程编号为”01″且课程成绩在60分以上的学生的学号和姓名;

select
    sid,cid,score
from sc
where cid='01' and score>60
40、查询选修“张三”老师所授课程的学生中,成绩最高的学生姓名及其成绩

select
    sc.sid,sname,cname,score
from sc
left join course
    on sc.cid=course.cid
left join teacher
    on course.tid=teacher.tid
left join student
    on sc.sid=student.sid
where tname='张三'
order by score desc
limit 1;
42、查询每门功课成绩最好的前两名

##感觉题目重复了
select
    cid,sid,rank1
from 
    (
        select
            cid
            ,sid
            ,rank() over(partition by cid order by score desc) as rank1
        from sc 
    )t
where rank1 <=2
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

select
    cid
    ,count(sid) as cnt
from sc
group by cid
having cnt>=5
order by count(sid) desc,cid
44、检索至少选修两门课程的学生学号

select
    sid
    ,count(cid)
from sc
group by sid
having count(cid)>=2
45、查询选修了全部课程的学生信息

#不太严谨,但实务中应该没问题,如需严谨见12题思路
select
    sid
    ,count(cid)
from sc
group by sid
having count(cid)=(select count(distinct cid) from sc)
46、查询各学生的年龄

select
    sid,sname,year(curdate())-year(sage) as sage
from student
47、查询本周过生日的学生

select
    sid,sname,sage
from student
where weekofyear(sage)=weekofyear(curdate())
48、查询下周过生日的学生

select 
    sid,sname,sage
from student
where weekofyear(sage) = weekofyear(date_add(curdate(),interval 1 week))
49、查询本月过生日的学生

select
    sid,sname,sage
from student
where month(sage) = month(curdate())
50、查询下月过生日的学生

select
    sid,sname,sage
from student
where month(date_sub(sage,interval 1 month)) = month(curdate())

题目及答案来源:(https://zhuanlan.zhihu.com/p/53302593)


 上一篇
数据科学速查手册 数据科学速查手册
Python基础入门 Python进阶知识 jupyter基础操作 Pandas基础知识 Numpy基础知识
2021-06-19
下一篇 
心理学研究方法笔记 心理学研究方法笔记
一、科学与常识 常识主要指在经验基础上形成的对客观世界的认识;科学则是在系统观察基础上形成的对客观世界的认识。 1、相似点 常识被定义为:在人们的实践基础上产生的概念以及对事物的一般性认识, 科学则是对常识的系统化和对常识的在控制基础上的
2021-06-19
// // //
  目录