版权声明：本文为博主原创文章，遵循 CC 4.0 BY-SA 版权协议，转载请附上原文出处链接和本声明。
根据流传着各种版本的Mysql 50题，做一份自己的解答跟进；
已经是第二遍再次回顾这部分题目，如有错误欢迎纠正～
另外十分感谢@启明星的指引 所提供的原文，附链接如下，题目摘自博主原文，答案为个人所写。
附第二弹题库链接：
SQL题库–45题版
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另附个人期间所使用工具
sqlfiddle在线工具
网页版方便个人在闲暇时间的小题目练习
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版权声明：本文为CSDN博主「启明星的指引」的原创文章，遵循 CC 4.0 BY-SA 版权协议，转载请附上原文出处链接及本声明。
原文链接：https://blog.csdn.net/fashion2014/article/details/78826299
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正文
建表
表名和字段
–1.学生表
Student(s_id,s_name,s_birth,s_sex) --学生编号,学生姓名, 出生年月,学生性别
–2.课程表
Course(c_id,c_name,t_id) – --课程编号, 课程名称, 教师编号
–3.教师表
Teacher(t_id,t_name) --教师编号,教师姓名
–4.成绩表
Score(s_id,c_id,s_score) --学生编号,课程编号,分数
--建表
--学生表
CREATE TABLE `Student`(
`s_id` VARCHAR(20),
`s_name` VARCHAR(20) NOT NULL DEFAULT '',
`s_birth` VARCHAR(20) NOT NULL DEFAULT '',
`s_sex` VARCHAR(10) NOT NULL DEFAULT '',
PRIMARY KEY(`s_id`)
);
--课程表
CREATE TABLE `Course`(
`c_id` VARCHAR(20),
`c_name` VARCHAR(20) NOT NULL DEFAULT '',
`t_id` VARCHAR(20) NOT NULL,
PRIMARY KEY(`c_id`)
);
--教师表
CREATE TABLE `Teacher`(
`t_id` VARCHAR(20),
`t_name` VARCHAR(20) NOT NULL DEFAULT '',
PRIMARY KEY(`t_id`)
);
--成绩表
CREATE TABLE `Score`(
`s_id` VARCHAR(20),
`c_id` VARCHAR(20),
`s_score` INT(3),
PRIMARY KEY(`s_id`,`c_id`)
);
--插入学生表测试数据
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');
--课程表测试数据
insert into Course values('01' , '语文' , '02');
insert into Course values('02' , '数学' , '01');
insert into Course values('03' , '英语' , '03');
--教师表测试数据
insert into Teacher values('01' , '张三');
insert into Teacher values('02' , '李四');
insert into Teacher values('03' , '王五');
--成绩表测试数据
insert into Score values('01' , '01' , 80);
insert into Score values('01' , '02' , 90);
insert into Score values('01' , '03' , 99);
insert into Score values('02' , '01' , 70);
insert into Score values('02' , '02' , 60);
insert into Score values('02' , '03' , 80);
insert into Score values('03' , '01' , 80);
insert into Score values('03' , '02' , 80);
insert into Score values('03' , '03' , 80);
insert into Score values('04' , '01' , 50);
insert into Score values('04' , '02' , 30);
insert into Score values('04' , '03' , 20);
insert into Score values('05' , '01' , 76);
insert into Score values('05' , '02' , 87);
insert into Score values('06' , '01' , 31);
insert into Score values('06' , '03' , 34);
insert into Score values('07' , '02' , 89);
insert into Score values('07' , '03' , 98);
题目及解答
-- 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 a.* , b.s_score from Student as a
join Score as b on a.s_id = b.s_id
join Score as c on b.s_id = c.s_id
where b.c_id = '01' and c.c_id = '02' and b.s_score > c.s_score
-- 2、查询"01"课程比"02"课程成绩低的学生的信息及课程分数
select a.* ,b.s_score from student as a
join score as b on a.s_id=b.s_id
join score as c on b.s_id=c.s_id
where b.c_id = '01' and c.c_id = '02' and b.s_score < c.s_score
-- 3、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 a.s_id, a.s_name , AVG(b.s_score) as avg_score from student as a
join score as b on a.s_id = b.s_id
GROUP BY a.s_id , a.s_name HAVING avg_score >=60
-- 4、查询平均成绩小于60分的同学的学生编号和学生姓名和平均成绩
-- (包括有成绩的和无成绩的)
select a.s_id ,a.s_name ,AVG(b.s_score)as avg_score from student as a
join score as b on a.s_id = b.s_id
GROUP BY a.s_id, a.s_name HAVING avg_score <60
union
select a.s_id,a.s_name,0 as avg_score from student as a
where a.s_id not in (select distinct s_id from score)
-- 5、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
select a.s_id,a.s_name,count(b.c_id) as sum_course,sum(b.s_score)
as sum_score from student as a
left join score b on a.s_id=b.s_id
GROUP BY a.s_id,a.s_name;
-- 6、查询"李"姓老师的数量
select count(t_name) from teacher
where t_name like '李%'
-- 7、查询学过"张三"老师授课的同学的信息
select * from student
where s_id in (select s_id from score
where c_id in (select c_id from course
where t_id in (select t_id from teacher
where t_name = '张三')))
-- 8、查询没学过"张三"老师授课的同学的信息
select * from student
where s_id in (select s_id from score
where c_id not in (select c_id from course
where t_id in (select t_id from teacher
where t_name = '张三')))
-- 9、查询学过编号为"01"并且也学过编号为"02"的课程的同学的信息
select * from student as a
join score as b on a.s_id=b.s_id and b.c_id = '01'
join score as c on c.s_id=b.s_id and c.c_id = '02'
-- 10、查询学过编号为"01"但是没有学过编号为"02"的课程的同学的信息
select * from student
where s_id in (select s_id from score where c_id = '01')
and s_id not in (select s_id from score where c_id ='02')
-- 11、查询没有学全所有课程的同学的信息
select a.* from student as a left join score as b on a.s_id=b.s_id
group by a.s_id having count(b.c_id) < (select count (*) from course)
--提示：使用left join保留student 表中一条数据，join会缺失（无分数）
-- 12、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 a.* from student as a
join score as b on a.s_id = b.s_id
where b.c_id in (select c_id from score where s_id ='01')
group by a.s_id
-- 13、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
-- 第二遍开始思路仍然是模糊的哈哈
select * from student
where s_id in
(select s_id from score group by s_id HAVING COUNT(s_id)=
-- 下面的语句是找到'01'同学学习的课程数
(select count (c_id) from score where s_id='01'))
and s_id not in
-- 下面的语句是找到学过‘01’同学没学过的课程，有哪些同学。并排除他们
(select s_id from score where c_id in (
-- 下面的语句是找到‘01’同学没学过的课程
select distinct c_id from score
where c_id not in (
-- 下面的语句是找出‘01’同学学习的课程
select c_id from score where s_id='01')
group by s_id))
-- 下面的条件是排除01同学
and s_id not in ('01')
-- 14、查询没学过"张三"老师讲授的任一门课程的学生姓名
select s_name from student
where s_id not in (select s_id from score
where c_id in (select c_id from course
where t_id in (select t_id from teacher
where t_name = '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
--个人解法1，错误方法，暂未发现错误，欢迎纠正～
select a.s_id,a.s_name,avg(b.s_score) as avg_score from student as a
join score as b on a.s_id = b.s_id
group by a.s_id
having (select count
(case when (b.s_score < 60 ) then 1 else 0 end )from score)>=2
select a.s_id,a.s_name,ROUND(AVG(b.s_score)) from
student a
left join score b on a.s_id = b.s_id
where a.s_id in(
select s_id from score where s_score<60 GROUP BY s_id having count(1)>=2)
GROUP BY a.s_id,a.s_name
-- 16、检索"01"课程分数小于60，按分数降序排列的学生信息
select * from student
where s_id in
(select s_id from score
where c_id = '01' and s_score <60
order by s_score desc)
-- 17、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 a.s_id,
(select s_score from score where a.s_id=s_id and c_id='01') as 语文,
(select s_score from score where a.s_id=s_id and c_id='02') as 数学,
(select s_score from score where a.s_id=s_id and c_id='03') as 英语,
avg(a.s_score) as avg_score from score as a
group by a.s_id
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，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
--原解法使用round四舍五入。也可以使用cast优化显示
select a.c_id,b.c_name,max(a.s_score),min(a.s_score),avg(a.s_score) as avg_score,
(sum (case when a.s_score >=60 then 1 else 0 end) / sum(case when a.s_score then 1 else 0 end)) as 及格率,
(sum (case when 70<a.s_score and a.s_score<80 then 1 else 0 end) / sum(case when a.s_score then 1 else 0 end)) as 中等率,
(sum (case when 80<a.s_score and a.s_score<90 then 1 else 0 end) / sum(case when a.s_score then 1 else 0 end)) as 优良率,
(sum (case when 90<= a.s_score then 1 else 0 end)/ sum (case when a.s_score then 1 else 0 end)) as 优秀率
from score as a join course as b on a.c_id=b.c_id
group by a.c_id
-- 19、按各科成绩进行排序，并显示排名
-- 筛选课程号‘01’的排序
select * from
(select t1.c_id,t1.s_score,
-- 选择2号score表中不重复的项目，条件为，2号score表中的分数大于1号score表并且课程号是‘01’，注释为rank
(select count(distinct t2.s_score) from score t2 where t2.s_score>=t1.s_score and t2.c_id='01') rank
-- 大条件限制为1号score表中课程号为‘01’的数据
FROM score t1 where t1.c_id='01'
order by t1.s_score desc) t1
union
-- 筛选课程号‘02’的排序
select * from (select t1.c_id,t1.s_score,
(select count(distinct t2.s_score) from score t2 where t2.s_score>=t1.s_score and t2.c_id='02') rank
FROM score t1 where t1.c_id='02'
order by t1.s_score desc) t2
union
-- 筛选课程号‘03’的排序
select * from (select
t1.c_id,
t1.s_score,
(select count(distinct t2.s_score) from score t2 where t2.s_score>=t1.s_score and t2.c_id='03') rank
FROM score t1 where t1.c_id='03'
order by t1.s_score desc) t3
-- ADD：还有一种全局变量赋值的解法，还没研究完毕 后续有机会补充
-- 20、查询学生的总成绩并进行排名
select s_id
@r := @r + 1 as i
@s := (case when @score =a.sum_score then @s else @r end) as rank
@score := a.sum_score as score
from(select s_id ,sum(s_score) as sum_score from score group by s_id order by sum_score desc) as a
(select @k:=0,@i:=0,@score:=0) as s
--sql fiddle 不支持，需mysql跑
-- 21、查询不同老师所教不同课程平均分从高到低显示
select c.t_id , b.c_id , avg(a.s_score) as avg_score from score as a
left join course as b on a.c_id = b.c_id
left join teacher as c on b.t_id = c.t_id
group by c.t_id
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 a.c_id ,a.s_score from score as a
where (select count(1) from score as b
where b.c_id=a.c_id and b.s_score >=a.s_score) =2
or (select count(1) from score as c
where c.c_id=a.c_id and c.s_score >=a.s_score) =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
-- 第二次做题的遗漏点是，忘记了筛选和join的匹配，遗失了几行数据
select a.c_id,'85-100',百分比1,'70-85',百分比2,'60-70',百分比3,'0-60',百分比4 from score as a
join (select c_id, sum(case when s_score>85 and s_score <=100 then 1 else 0 end) as '85-100',
round ('85-100'/count(*),2) as 百分比1 from score group by c_id) as b on a.c_id=b.c_id
join (select c_id, sum(case when s_score>70 and s_score <=85 then 1 else 0 end) as '70-85',
round ('70-85'/count(*),2) as 百分比2 from score group by c_id) as c on a.c_id=c.c_id
join (select c_id, sum(case when s_score>60 and s_score <=70 then 1 else 0 end) as '60-70',
round ('60-70'/count(*),2) as 百分比3 from score group by c_id) as d on a.c_id=d.c_id
join (select c_id, sum(case when s_score>=0 and s_score <=60 then 1 else 0 end) as '0-60',
round ('0-60'/count(*),2) as 百分比4 from score group by c_id) as e on a.c_id=e.c_id
group by a.c_id
-- 24、查询学生平均成绩及其名次
-- 全局变量解法典型例题
select a.s_id,
@i:=@i+1 as '不保留空缺排名',
@k:=(case when @avg_score=a.avg_s then @k else @i end) as '保留空缺排名',
@avg_score:=avg_s as '平均分'
from (select s_id,ROUND(AVG(s_score),2) as avg_s from score GROUP BY s_id ORDER BY avg_s DESC)a,(select @avg_score:=0,@i:=0,@k:=0)b;
-- 25、查询各科成绩前三名的记录
-- 1.选出b表比a表成绩大的所有组
-- 2.选出比当前id成绩大的 小于三个的
select * from score as a join score as b
on a.s_id=b.s_id and b.s_score>a.s_score
group by a.s_id
having count(b.s_score)<3
order by b.s_score desc
-- 26、查询每门课程被选修的学生数
select c_id,count(s_id) from score a GROUP BY c_id
-- 27、查询出只有两门课程的全部学生的学号和姓名
select a.s_id,b.s_name from score as a
join student as b on a.s_id=b.s_id
group by a.s_id
having count(s_score)=2
-- 28、查询男生、女生人数
select s_sex, count(*) from student
group by s_sex
-- 29、查询名字中含有"风"字的学生信息
select * from student
where s_name like '%风%'
-- 30、查询同名同性学生名单，并统计同名人数
select *,count(*) from student as a
join student as b on a.s_name=b.s_name and a.s_sex=b.s_sex
group by a.s_name,a.s_sex
-- 31、查询1990年出生的学生名单
select s_name from student
where s_birth like '1990%'
-- 32、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 c_id,round(avg(s_score),2) as avg_score from score
group by c_id
order by avg(s_score) desc,c_id asc
-- 33、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 a.s_id,a.s_name,avg(b.s_score) as avg_score from student as a
join score as b on a.s_id=b.s_id
group by a.s_id
having avg_score>=85
-- 34、查询课程名称为"数学"，且分数低于60的学生姓名和分数
select a.s_name ,b.s_score from student as a
join score as b on a.s_id = b.s_id
where b.c_id in (select c_id from course
where c_name='数学') and b.s_score<60
-- 35、查询所有学生的课程及分数情况；
select a.c_id ,
sum(case when b.c_name='语文' then a.s_score else 0 end) as '语文',
sum(case when b.c_name='数学' then a.s_score else 0 end) as '数学',
sum(case when b.c_name='英语' then a.s_score else 0 end) as '英语',
sum(a.s_score)as '总分'
from score as a join course as b on a.c_id=b.c_id
-- 36、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 c.s_name ,b.c_name ,a.s_score from score as a
join course as b on a.c_id=b.c_id
join student as c on a.s_id=c.s_id
where a.s_score>=70
-- 37、查询不及格的课程
select a.c_id,b.c_name,a.s_score from score a
left join course b on a.c_id = b.c_id
where a.s_score<60
-- 38、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 a.s_id ,a.s_name from student as a
join score as b on a.s_id=b.s_id
where b.c_id='01' and b.s_score >80
-- 39、求每门课程的学生人数
select a.c_id ,b.c_name,count(*) from score as a
join course as b on a.c_id=b.c_id
group by a.c_id
-- 40、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 a.*,max(b.s_score) from student as a
join score as b on a.s_id=b.s_id
where c_id in (select c_id from course
where t_id in (select t_id from teacher
where t_name = '张三'))
-- 41、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 distinct a.s_id,a.c_id,a.s_score from score as a
join score as b on a.c_id != b.c_id and a.s_score =b.s_score
-- 42、查询每门功成绩最好的前两名
select a.s_id ,a.c_id ,a.s_score from score as a
where (select count(1) from score as b
where b.c_id=a.c_id and b.s_score >=a.s_score ) <=2
order by a.c_id
-- 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 c_id ,count(*) as total from score
group by c_id
having total>5
order by total desc ,c_id asc
-- 44、检索至少选修两门课程的学生学号
select s_id,count(s_score)as tol from score
group by s_id
having tol >=2
-- 45、查询选修了全部课程的学生信息
select * from student
where s_id in ( select s_id from score group by s_id having count(*) =(select count(*) from course))
-- 46、查询各学生的年龄
-- 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 s_birth,(DATE_FORMAT(NOW(),'%Y')-DATE_FORMAT(s_birth,'%Y') -
(case when DATE_FORMAT(NOW(),'%m%d')>DATE_FORMAT(s_birth,'%m%d') then 0 else 1 end))
as age from student
-- 47、查询本周过生日的学生
select * from student where WEEK(DATE_FORMAT(NOW(),'%Y%m%d'))=WEEK(s_birth)
-- 48、查询下周过生日的学生
select * from student where WEEK(DATE_FORMAT(NOW(),'%Y%m%d'))+1 =WEEK(s_birth)
-- 49、查询本月过生日的学生
select * from student where MONTH(DATE_FORMAT(NOW(),'%Y%m%d')) =MONTH(s_birth)
-- 50、查询下月过生日的学生
select * from student where MONTH(DATE_FORMAT(NOW(),'%Y%m%d'))+1 =MONTH(s_birth)
Finish~~~