我的私藏SQL练习题

作者: 2453cf172ab4 | 来源:发表于2017-04-05 19:31 被阅读1116次

    测试数据

    环境:Mysql

    测试数据集,总共四张表,以及一些初始化数据,模拟一个小的场景,练习使用。

    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' , '1990-05-20' , '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');
    insert into Student values('09' , '孙吴昊' , '1990-01-20' , '女');
    insert into Student values('10' , '赵雷' , '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');
    insert into Course values('04' , '英语' , '01');
    
    
    create table Teacher(
    Tid varchar(10),
    Tname varchar(10)
    );
    
    insert into Teacher values('01' , N'张三');
    insert into Teacher values('02' , N'李四');
    insert into Teacher values('03' , N'王五');
    insert into Teacher values('04' , N'汪二蛋');
    
    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);
    insert into SC values('08' , '03' , 98);
    insert into SC values('09' , '03' , 98);
    insert into SC values('10' , '04' , 59);
    
    

    试题

    题1

    查询"01"课程比"02"课程成绩高的学生的信息及课程分数。

    答案1

    查询同时存在"01"课程和"02"课程的情况。

    
    select a.*, b.score, c.score
    from student a, sc b,sc c
    where a.sid = b.sid and a.sid = c.sid 
    and b.cid ='01' and c.cid = '02'
    and b.score > c.score;
    
    

    答案2

    查询:同时存在"01"课程和"02"课程的情况和存在"01"课程但可能不存在"02"课程的情况(不存在时显示为null)。

    
    select a.*, b.score, c.score
    from student a 
    left join sc b on a.sid = b.sid and b.cid = '01'
    left join sc c on a.sid = c.sid and c.cid = '02'
    where b.score > ifnull(c.score,0);
    
    

    题2

    查询平均成绩大于等于60分的同学的学生编号、学生姓名和平均成绩。

    答案1

    select a.sid, a.sname, avg(sc.score) avg_score
    from student a, sc
    where a.sid = sc.sid
    group by a.sid, a.sname
    having avg_score >= 60
    order by a.sid;
    

    答案2

    使用cast转换精度。

    select a.sid, a.sname, avg(cast(sc.score as decimal(18,2))) avg_score
    from student a, sc
    where a.sid = sc.sid
    group by a.sid, a.sname
    having avg_score >= 60
    order by a.sid;
    

    题3

    查询平均成绩小于60分的同学的学生编号、学生姓名和平均成绩(包含无成绩的)。

    答案

    select a.sid, a.sname, avg(ifnull(sc.score,0)) avg_score
    from student a left join sc
    on a.sid = sc.sid
    group by a.sname, a.sid
    having avg_score < 60
    order by a.sid;
    

    题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。

    答案1

    查询所有有成绩的SQL。

    select a.sid, a.sname, count(b.cid), sum(b.score)
    from student a inner join sc b
    on a.sid = b.sid
    group by a.sid, a.sname
    order by a.sid;
    

    答案2

    查询所有(包括有成绩和无成绩)的SQL。

    select a.sid, a.sname, count(b.cid), sum(b.score)
    from student a left join sc b
    on a.sid = b.sid
    group by a.sid, a.sname
    order by a.sid;
    

    题5

    查询"李"姓老师的数量。

    答案

    select count(tname)
    from teacher
    where tname like '李%';
    

    题6

    统计课程分01最高的前三名。

    答案

    select student.sname, sc.score, sc.cid
    from student join sc on student.sid = sc.sid 
    where sc.cid='01'
    order by sc.score desc limit 3;
    

    题7

    统计课程分最高的前三名,不分课程(分数的前三,人数可能大于三)

    答案1

    使用join。

    select a.sname, a.score,  a.cid
    from 
    (select Student.sname, sc.score, sc.cid
    from Student 
    join SC on Student.Sid = SC.Sid) as a 
    join 
    (select distinct sc.score
    from SC
    order by score 
    desc limit 3) as b
    on a.score = b.score;
    

    答案2

    使用in。

    select a.sname, a.score,  a.cid
    from 
    (select Student.sname, sc.score, sc.cid
    from Student 
    join SC on Student.Sid = SC.Sid) as a 
    where a.score in
    (select distinct sc.score from sc
    order by sc.score
    desc limit 3
    );
    

    题8

    统计分数出现次数前三的分数

    答案

    select score, count(*)
    from SC
    group by score
    order by count(*) 
    desc limit 3; 
    

    题9

    统计每个学生选课总数。

    答案

    select a.sid, a.sname, count(b.cid) 
    from student a left join sc b 
    on a.sid=b.sid
    group by a.sid,a.sname;
    

    题10

    每个老师教多少学生(一个教师给某个学生教两门课,计入两次)

    答案

    select t.tid, t.tname, a.countstu
    from
    (select c.tid as ctid, count(sc.cid) as countstu
    from course c inner join sc
    on c.cid = sc.cid
    group by c.tid) a
    right join teacher t
    on t.tid = a.ctid;
    

    2015-10-28 23:14:54 于bh xzl


    作者:dantezhao |简书 | CSDN | GITHUB

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    个人主页:http://www.jianshu.com/u/2453cf172ab4
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      • 9566c7cbfc5d:使用cast转换精度,cast是不是应该嵌套在avg函数外面,使结果值保留两位:cast( avg(sc.score) as decimal(18,2))

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