经典SQL50题，做完这些sql就入门了~

2023年12月31日 165次阅读来源: 木公鼠跪鱼

SQL50题

网上看的50道SQL训练题，试着做了一下，难度依次递增。欢迎一起学习的朋友进行讨论指出错误~

数据表介绍

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');

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');

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');

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

查询” 01 “课程比” 02 “课程成绩高的学生的信息及课程分数

select * from (
select s1.SId,s1.score class1,s2.score class2 from test.sc s1 
left join test.sc s2 on s1.SId = s2.SId
where s1.CId = "01" and s2.CId = "02" and s1.score > s2.score
)r left join test.student s on r.SId = s.SId;

查询同时存在” 01 “课程和” 02 “课程的情况

select s1.SID,s1.score class1,s2.score class2 from test.sc s1 
left join test.sc s2 on s1.SId = s2.SId 
where s1.CId = "01" and s2.CId = "02"

select * from 
 (select * from test.sc  where CId = "01")  t1,
 (select * from test.sc  where CId = "02")  t2
where  t1.SId = t2.SId;

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 )

select * from 
(select * from test.sc  where CId = "01")  t1
left join 
(select * from test.sc  where CId = "02")  t2
on  t1.SId = t2.SId;

查询不存在” 01 “课程但存在” 02 “课程的情况

select s1.* from test.sc s1
join test.sc s2 on s1.SId = s2.SId 
where s1.SId not in  (select SId from test.sc s where s.CId = "01")
AND s2.CId = "02";

和网上有些看到的稍微不太一样，我认为不存在01，存在02，是筛选出这个学生的SId，该学生其他课程应该也应展示出来。网上常见其他答案参考如下，仅展示了有02课程：

select *
from sc
where sc.SId not in (select SId from sc where sc.CId='01')
and  sc.CId='02';

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 r.*,s.Sname from 
(SELECT SId,AVG(score) FROM test.sc 
group by SId  having avg(score) > 60) r
left join test.student s on s.SId = r.SId;

select c.SId, s.Sname, avg(c.score)
from test.sc c inner join test.student s on c.SId=s.SId
group by c.SId
having avg(c.score)>=60;

查询在 SC 表存在成绩的学生信息

SELECT distinct t.* FROM test.sc s ,test.student t
where s.SId = t.SId;

查询所有同学的学生编号、学生姓名、选课总数、所有课程的成绩总和

SELECT t.SId,t.Sname,r.ccount 课程数量,r.csum 课程总分 FROM test.student t ,
(select s.SId,count(s.CId) ccount,sum(s.score) csum from test.sc s 
group by s.SId) r where t.SId = r.SId;

要显示没选课的学生(显示为NULL)，使用join:

select s.SId, s.Sname, count(c.CId), sum(score)
from test.student s left join test.sc c on s.SId=c.SId
group by c.SId

查有成绩的学生信息

select *
from student
where EXISTS(select * from sc where student.SId=sc.SId)

SQL查询中in和exists的区别分析

查询「李」姓老师的数量

SELECT count(Tid) FROM test.teacher 
where Tname like "李%";

查询学过「张三」老师授课的同学的信息

select distinct st.* from 
(
    select s.* from test.sc s,
	(
		SELECT c.CId FROM test.course c ,
		(SELECT TId FROM test.teacher  where Tname = "张三") t
		where c.TId = t.TId
	) cid
	where s.CId = cid.CId
)r
join test.student st on st.SId = r.SId;

上面自己写的，一张一张表查询的，效率可能较低。下面是其他人写的，

select distinct s.*
from test.student s
inner join test.sc sc on s.SId=sc.SId
inner join test.course c on sc.CId=c.CId
inner join test.teacher t on c.TId=t.TId
where t.Tname='张三'

查询没有学全所有课程的同学的信息

SELECT * FROM test.student t
where t.SId not in 
(
    select s.SId  from test.sc s 
    group by s.SId 
    having count(s.CId)>=3
);

其实这里3的来源是自己数出来有3个，实际应该也是查询出数量有多少，3可以替换成(select count(cid) from course)

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 s.SId,t.* FROM test.sc s , test.student t
where CId in 
(select CId from test.sc where sc.SId='01')
and s.SId = t.SId and s.SId !="01"

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  distinct t.SId ,t.Sname,t.Sage,t.Ssex
FROM test.sc s ,test.student t,(select CId from test.sc where sc.SId='01') s1
where s.SId = t.SId and t.SId!='01' and s1.CId =  s.CId
group by t.SId
having count(S.CId) > 2

查询没学过”张三”老师讲授的任一门课程的学生姓名

select * from test.student s
where s.SId not in 
(
	select sc.SId FROM test.sc as sc
	join
	(
		SELECT c.CId FROM test.course c,test.teacher t
		where c.TId = t.TId and t.Tname = "张三" 
	)r
	on sc.CId = r.CId
);

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 s.SId,s.Sname FROM test.student s
join 
(
 SELECT sc.SId,avg(score) FROM test.sc sc
 where sc.score < 60
 group by sc.SId 
 having count(sc.CId)>1
)r
on s.SId = r.SId ;

其他比我简便的做法

select s.SId, s.Sname, avg(sc1.score)
from (select * from test.sc where score<60) sc1, test.student s
where sc1.SId=s.SId
group by sc1.SId
having count(sc1.CId)>1;

检索” 01 “课程分数小于 60，按分数降序排列的学生信息

SELECT * FROM test.student sc,
(
 select s.SId,s.score from test.sc s 
 where s.CId = "01" and s.score < 60
 )r
where sc.SId = r.SId 
order by r.score desc;

我比较习惯有逻辑的写看的比较清晰，网上较多是比较精炼的如下：

select student.*
from student,sc
where sc.CId ='01'
and   sc.score<60
and   student.SId=sc.SId

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
    test.sc sc2,
    (SELECT 
        sc.SId, AVG(sc.score) average_score
    FROM
        test.sc sc
    GROUP BY sc.SId) r
WHERE
    r.SId = sc2.SId
ORDER BY r.average_score DESC;

突然发现了workbench的一个快捷键可以美化代码，哈哈哈，终于可以告别丑陋的书写了。

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	课程人数	最高分	最低分	平均分	及格率	中等率	优良率	优秀率
CId	Cname	number	max	min	mean	>=60	70-80	80-90	90-100

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 
    sc.CId,
    c.Cname,
    COUNT(DISTINCT sc.SId) 课程人数,
    AVG(sc.score) 平均分,
    MAX(sc.score) 最高分,
    MIN(sc.score) 最低分,
    COUNT(CASE
        WHEN sc.score >= 60 THEN 1
        ELSE NULL
    END) / COUNT(DISTINCT sc.SId) 及格率,
    COUNT(CASE
        WHEN sc.score >= 70 AND sc.score < 80 THEN 1
        ELSE NULL
    END) / COUNT(DISTINCT sc.SId) 中等率,
    COUNT(CASE
        WHEN sc.score >= 80 AND sc.score < 90 THEN 1
        ELSE NULL
    END) / COUNT(DISTINCT sc.SId) 优良率,
    COUNT(CASE
        WHEN sc.score >= 80 AND sc.score < 90 THEN 1
        ELSE NULL
    END) / COUNT(DISTINCT sc.SId) 优秀率
FROM
    test.sc sc,
    test.course c
WHERE
    c.CId = sc.CId
GROUP BY CId
ORDER BY 课程人数 DESC , sc.CId;

按各科成绩进行排序，并显示排名， Score 重复时保留名次空缺

SELECT 
    a.cid, a.sid, a.score,COUNT(b.score) + 1 AS Srank
FROM
    test.sc AS a
        LEFT JOIN
    teSt.sc AS b ON  a.score < b.score AND a.CId = b.CId
GROUP BY a.CId , a.SId , a.score
ORDER BY a.CId , Srank ASC;

这题有些难度没有做出来，参考其他人的方法，有点trick。

按各科成绩进行排序，并显示排名， Score 重复时合并名次

SELECT 
    sc.SId, sc.CId, sc.score, tp.ranks
FROM
    test.sc
    LEFT JOIN
        (SELECT 
            SId,
            CId,
            (SELECT 
                COUNT(DISTINCT sc2.score) + 1
                FROM
                    test.sc sc2
                WHERE
                    sc1.CId = sc2.CId
                AND sc2.score > sc1.score) ranks
         FROM test.sc sc1) tp 
    ON sc.SId = tp.SId AND sc.CId = tp.CId
ORDER BY sc.CId , ranks

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 t1.*,@currank:= @currank+1 as rank
from (select sc.SId, sum(score)
from sc
GROUP BY sc.SId 
ORDER BY sum(score) desc) as t1,(select @currank:=0) as t

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 t1.*, case when @fontscore=t1.sumscore then @currank  when @fontscore:=t1.sumscore  then @currank:=@currank+1  end as rank
from (select sc.SId, sum(score) as sumscore
from sc
GROUP BY sc.SId 
ORDER BY sum(score) desc) as t1,(select @currank:=0,@fontscore:=null) as t

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 
    c.CId 课程编号,
    c.Cname 课程名称,
    COUNT(CASE
            WHEN sc.score >= 85 THEN 1
            ELSE NULL
         END) AS '[100,85]人数',
    ROUND(COUNT(CASE
                    WHEN sc.score >= 85 THEN 1
                    ELSE NULL
                END) * 1.0 / COUNT(sc.score),
            2) AS '[100,85]占比',
    COUNT(CASE
            WHEN sc.score >= 70 AND sc.score < 85 THEN 1
            ELSE NULL
         END) AS '(85,70]人数',
    ROUND(COUNT(CASE
                    WHEN sc.score >= 70 AND sc.score < 85 THEN 1
                    ELSE NULL
                END) * 1.0 / COUNT(sc.score),
         2) AS '(85,70]占比',
    COUNT(CASE
            WHEN sc.score >= 60 AND sc.score < 70 THEN 1
            ELSE NULL
         END) AS '(70,60]人数',
    ROUND(COUNT(CASE
                    WHEN sc.score >= 60 AND sc.score < 70 THEN 1
                    ELSE NULL
                 END) * 1.0 / COUNT(sc.score),
         2) AS '(70,60]占比',
    COUNT(CASE
            WHEN sc.score < 60 THEN 1
            ELSE NULL
          END) AS '(60,0]人数',
    ROUND(COUNT(CASE
                    WHEN sc.score < 60 THEN 1
                    ELSE NULL
                END) * 1.0 / COUNT(sc.score),
         2) AS '(60,0]占比'
FROM
    Course c
    LEFT JOIN sc 
    ON c.CId = sc.CId
GROUP BY c.CId

查询各科成绩前三名的记录

select *
from sc  
where  (select count(*) from sc as a where sc.CId =a.CId and  sc.score <a.score )<3
ORDER BY CId asc,sc.score desc

这个就是前三名转化为若大于此成绩的数量少于3即为前三名。

查询每门课程被选修的学生数

SELECT 
    CId, COUNT(SId)
FROM
    test.sc
GROUP BY CId
;

查询出只选修两门课程的学生学号和姓名

SELECT 
    s.SId, s.Sname
FROM
    test.student s,
    test.sc sc
WHERE
    s.SId = sc.SId
GROUP BY s.SId
HAVING COUNT(sc.CId) = 2
;

查询男生、女生人数

SELECT 
    Ssex, COUNT(SId)
FROM
    test.student
GROUP BY Ssex
;

查询名字中含有「风」字的学生信息

SELECT 
    *
FROM
    test.student
WHERE
    Sname LIKE '%风%'
;

查询同名学生名单，并统计同名人数

SELECT 
    Sname, COUNT(Sname)
FROM
    test.student
GROUP BY Sname
HAVING COUNT(Sname) > 1;

查询 1990 年出生的学生名单

SELECT 
    *
FROM
    test.student
WHERE
    YEAR(student.Sage) = 1990;

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) avg_score
FROM
    test.sc
GROUP BY CId
ORDER BY avg_score DESC , CId;

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 
    sc.SId, s.Sname, AVG(sc.score) avg_score
FROM
    test.sc sc,
    test.student s
WHERE
    sc.SId = s.SId
GROUP BY sc.SId
HAVING SUM(score) >= 85;

查询课程名称为「数学」，且分数低于 60 的学生姓名和分数

SELECT 
    c.Cname, s.Sname, sc.score
FROM
    test.sc sc,
    test.course c,
    test.student s
WHERE
    sc.CId = c.CId 
    AND c.Cname = '数学'
    AND sc.SId = s.SId
    AND sc.score < 60;

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 
    s.SId, s.Sname, sc.CId, sc.score
FROM
    test.student s
LEFT JOIN
    test.sc sc 
ON 
    sc.SId = s.SId;

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 
    s.Sname, c.Cname, sc.score
FROM
    test.sc sc,
    test.student s,
    test.course c
WHERE
    sc.score > 70 
    AND sc.SId = s.SId
    AND c.CId = sc.CId;

查询存在不及格的课程

SELECT DISTINCT
    c.Cname
FROM
    test.sc sc,
    test.course c
WHERE
    sc.score < 60

查询课程编号为 01 且课程成绩在 80 分及以上的学生的学号和姓名

SELECT DISTINCT
    s.SId, s.Sname
FROM
    test.sc sc,
    test.student s
WHERE
    sc.score >= 80 
    AND sc.CId = '01'
    AND sc.SId = s.SId;

求每门课程的学生人数

SELECT 
    sc.CId, COUNT(*) AS 学生人数
FROM
    test.sc
GROUP BY sc.CId;

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 DISTINCT
    s.*, sc.score
FROM
    test.teacher t,
    test.sc sc,
    test.student s,
    test.course c
WHERE
    t.TId = c.TId 
    AND t.Tname = '张三'
    AND c.CId = sc.CId
    AND s.SId = sc.SId
ORDER BY sc.score DESC
LIMIT 1;

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 
    student.*, sc.score, sc.cid
FROM
    student,
    teacher,
    course,
    sc
WHERE
    teacher.tid = course.tid
    AND sc.sid = student.sid
    AND sc.cid = course.cid
    AND teacher.tname = '张三'
    AND sc.score = (
        SELECT MAX(sc.score)
        FROM
            sc,
            student,
            teacher,
            course
        WHERE
            teacher.tid = course.tid
            AND sc.sid = student.sid
            AND sc.cid = course.cid
            AND teacher.tname = '张三');

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 
    sc1.SId, sc1.CId, sc1.score
FROM
    test.sc sc1
JOIN
    test.sc sc2 
ON  sc1.SId = sc2.SId
    AND sc1.score = sc2.score
    AND sc1.CId != sc2.CId
GROUP BY sc1.CId , sc1.SId;

查询每门功成绩最好的前两名

SELECT 
    a.CId, a.SId, a.score
FROM
    test.sc a
LEFT JOIN
    test.sc b 
ON  
    a.CId = b.CId 
    AND a.score < b.score
GROUP BY a.CId , a.SId
HAVING COUNT(a.CId) < 2
ORDER BY CId , score DESC

同19题类似的思路

统计每门课程的学生选修人数（超过 5 人的课程才统计）。

SELECT a.CId, count(a.SId) FROM test.sc a
GROUP BY a.CId 
HAVING COUNT(a.SId) > 5;

检索至少选修两门课程的学生学号

SELECT 
    sc.SId, COUNT(CId) AS cn
FROM
    test.sc
GROUP BY SId
HAVING cn > 1;

查询选修了全部课程的学生信息

SELECT 
    sc.SId, COUNT(CId) AS cn
FROM
    test.sc
GROUP BY SId
HAVING cn = (SELECT COUNT(cid) FROM test.course);

查询各学生的年龄，只按年份来算
同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,student.Sage,CURDATE())  age
FROM
    test.student；

mysql的timestampdiff函数很好用，满足以上需要
42. 查询本周过生日的学生

SELECT 
    *
FROM
    test.student
WHERE
    WEEKOFYEAR(student.Sage) = WEEKOFYEAR(CURDATE());

查询下周过生日的学生

SELECT 
    *
FROM
    test.student
WHERE
    WEEKOFYEAR(student.Sage) - WEEKOFYEAR(CURDATE()) = 1;

查询本月过生日的学生

SELECT 
    *
FROM
    test.student
WHERE
    MONTH(student.Sage) = MONTH(CURDATE());

查询下月过生日的学生

SELECT 
    *
FROM
    test.student
WHERE
    MONTH(student.Sage) = MONTH(CURDATE())+1;

终于写完了，前9题就是基本的操作，并且难度逐渐增加的。而之后主要就是熟悉各种函数，做完后除了对变量还不是很熟悉以外，其他感觉还不错。果然实践做题才是学习sql最好的方式。

    原文作者：木公鼠跪鱼
    原文地址: https://blog.csdn.net/weixin_39337018/article/details/83177627
    本文转自网络文章，转载此文章仅为分享知识，如有侵权，请联系博主进行删除。