MySQl经典50题，全部都会做就是大佬了！

程序员文章站 2022-06-16 08:35:24

MySQL经典50题，全部都会做就是大佬了！文章目录MySQL经典50题，全部都会做就是大佬了！创建表结构建表语句添加数据表关系题目参考答案1.查询" 01 "课程⽐" 02 "课程成绩⾼的学⽣的信息及课程分数2.查询同时存在" 01 "课程和" 02 "课程的情况3.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 )4.查询不存在" 01 "课程但存在" 02 "课程的情况5.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6.查询在 SC 表存在...

MySQL经典50题，全部都会做就是大佬了！

文章目录

MySQL经典50题，全部都会做就是大佬了！

创建表结构

建表语句

-- 学⽣表 Student
create table Student(
    SId varchar(10),
    Sname varchar(10),
    Sage datetime,
    Ssex varchar(10)
);

-- 科⽬表 Course
create table Course(
    CId varchar(10),
    Cname nvarchar(10),
    TId varchar(10)
);

-- 教师表 Teacher
create table Teacher(
    TId varchar(10),
    Tname varchar(10)
);

-- 成绩表 SC
create table SC(
    SId varchar(10),
    CId varchar(10),
    score decimal(18,1)
);

添加数据

-- 学⽣表 Student
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' , '孙⻛' , '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');

-- 科⽬表 Course
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
insert into Teacher values('01' , '张三');
insert into Teacher values('02' , '李四');
insert into Teacher values('03' , '王五');

-- 成绩表 
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);

表关系

稍微写几道题就熟练了！

MySQl经典50题，全部都会做就是大佬了！

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

2.查询同时存在" 01 “课程和” 02 "课程的情况

3.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 )

4.查询不存在" 01 “课程但存在” 02 "课程的情况

5.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

6.查询在 SC 表存在成绩的学⽣信息

7.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 )

8.查询「李」姓⽼师的数量

9.查询学过「张三」⽼师授课的同学的信息

10.查询没有学全所有课程的同学的信息

11.查询⾄少有⼀⻔课与学号为" 01 "的同学所学相同的同学的信息

12.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

13.查询没学过"张三"⽼师讲授的任⼀⻔课程的学⽣姓名

14.查询两⻔及其以上不及格课程的同学的学号，姓名及其平均成绩

15.检索" 01 "课程分数⼩于 60，按分数降序排列的学⽣信息

16.按平均成绩从⾼到低显示所有学⽣的所有课的成绩以及平均成绩

17.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 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

18.按各科平均成绩进⾏排序，并显示排名， Score 重复时保留名次空缺

19.按各科平均成绩进⾏排序，并显示排名， Score 重复时不保留名次空缺

20.查询学⽣的总成绩，并进⾏排名，总分重复时保留名次空缺

21.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

22.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

23.查询各科成绩前三名的记录

24.查询每⻔课程被选修的学⽣数

25.查询出只选修两⻔课程的学⽣学号和姓名

26.查询男⽣、⼥⽣⼈数27.查询名字中含有「⻛」字的学⽣信息

28.查询同名同性学⽣名单，并统计同名⼈数

29.查询 1990 年出⽣的学⽣名单

30.查询每⻔课程的平均成绩，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

31.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

32.查询课程名称为「数学」，且分数低于 60 的学⽣姓名和分数

33.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）

34.查询任何⼀⻔课程成绩在 70 分以上的姓名、课程名称和分数

35.查询不及格的课程

36.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

37.求每⻔课程的学⽣⼈数

38.成绩不重复，查询选修「张三」⽼师所授课程的学⽣中，成绩最⾼的学⽣信及其成绩

39.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

40.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

41.查询每⻔课程成绩最好的前两名

42.统计每⻔课程的学⽣选修⼈数（超过 5 ⼈的课程才统计）。

43.检索⾄少选修两⻔课程的学⽣学号

44.查询选修了全部课程的学⽣信息

45.查询各学⽣的年龄，只按年份来算

46.按照出⽣⽇期来算，当前⽉⽇ < 出⽣年⽉的⽉⽇则，年龄减⼀TIMESTAMPDIFF() 从⽇期时间表达式中减去间隔

47.查询本周过⽣⽇的学⽣

返回⽇期从范围内的数字⽇历星期1到5348.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 "课程成绩⾼的学⽣的信息及课程分数


分析：我们这里使用到了学生信息，课程分数，所以我们需要关联三张表。
先分别查询01和02课程的学员的id和分数

select sid,score from sc where cid='01';
select sid,score from sc where cid='02';

+------+-------+                
| sid  | score |                
+------+-------+                
| 01   |  80.0 |                
| 02   |  70.0 |                
| 03   |  80.0 |                
| 04   |  50.0 |                
| 05   |  76.0 |                
| 06   |  31.0 |                
+------+-------+                
6 rows in set (0.00 sec)        
                                
+------+-------+                
| sid  | score |                
+------+-------+                
| 01   |  90.0 |                
| 02   |  60.0 |                
| 03   |  80.0 |                
| 04   |  30.0 |                
| 05   |  87.0 |                
| 07   |  89.0 |                
+------+-------+                
6 rows in set (0.00 sec)        

对比结果发现，两个结果中有些sid是不对应的
因此可以对两个结果做join联结，同时sid作为关联条件
并且01乘积大于02

select s1.sid,s1.score from
(select sid,score from sc where cid='01') as s1
join
(select sid,score from sc where cid='02') as s2
on s1.sid=s2.sid
where s1.score > s2.score;

结果
+------+-------+
| sid  | score |
+------+-------+
| 02   |  70.0 |
| 04   |  50.0 |
+------+-------+

题目要求是学生信息，而不是学生id，所以进一步查询

select stu.* ,s.score
from student as stu
right join
(
	select s1.sid,s1.score from
	(select sid,score from sc where cid='01') as s1
	join
	(select sid,score from sc where cid='02') as s2
	on s1.sid=s2.sid
	where s1.score > s2.score
) as s
on stu.sid = s.sid;

结果
+------+--------+---------------------+------+-------+
| SId  | Sname  | Sage                | Ssex | score |
+------+--------+---------------------+------+-------+
| 02   | 钱电   | 1990-12-21 00:00:00 | 男   |  70.0 |
| 04   | 李云   | 1990-12-06 00:00:00 | 男   |  50.0 |
+------+--------+---------------------+------+-------+

有点复杂我们简化一下

select student.*, s1.score, s2.score
from student
 join sc s1 on s1.sid=student.sid and s1.cid='01'
 join sc s2 on s2.sid=student.sid and s2.cid='02'
where s1.score>s2.score

结果一致
+------+--------+---------------------+------+-------+
| SId  | Sname  | Sage                | Ssex | score |
+------+--------+---------------------+------+-------+
| 02   | 钱电   | 1990-12-21 00:00:00 | 男   |  70.0 |
| 04   | 李云   | 1990-12-06 00:00:00 | 男   |  50.0 |
+------+--------+---------------------+------+-------+

2.查询同时存在" 01 “课程和” 02 "课程的情况


select t1.SId ,t1.score '01' , t2.score '02' 
from (select * from sc WHERE sc.CId='01')as t1 
inner join (select * from sc WHERE sc.CId='02')as t2 ON t1.SId=t2.SId


+------+------+------+    
| SId  | 01   | 02   |    
+------+------+------+    
| 01   | 80.0 | 90.0 |    
| 02   | 70.0 | 60.0 |    
| 03   | 80.0 | 80.0 |    
| 04   | 50.0 | 30.0 |    
| 05   | 76.0 | 87.0 |    
+------+------+------+

3.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 t1.SId ,t1.score '01' , t2.score '02' 
from (select * from sc WHERE sc.CId='01')as t1 
LEFT JOIN (select * from sc WHERE sc.CId='02')as t2 ON t1.SId=t2.SId

+------+------+------+
| SId  | 01   | 02   |
+------+------+------+
| 01   | 80.0 | 90.0 |
| 02   | 70.0 | 60.0 |
| 03   | 80.0 | 80.0 |
| 04   | 50.0 | 30.0 |
| 05   | 76.0 | 87.0 |
| 06   | 31.0 | NULL |
+------+------+------+

4.查询不存在" 01 “课程但存在” 02 "课程的情况

select t1.SId ,t2.score '01' , t1.score '02' 
from (select * from sc WHERE sc.CId='02')as t1 
LEFT JOIN (select * from sc WHERE sc.CId='01')as t2 ON t1.SId=t2.SId

+------+------+------+
| SId  | 01   | 02   |
+------+------+------+
| 01   | 80.0 | 90.0 |
| 02   | 70.0 | 60.0 |
| 03   | 80.0 | 80.0 |
| 04   | 50.0 | 30.0 |
| 05   | 76.0 | 87.0 |
| 07   | NULL | 89.0 |
+------+------+------+

5.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 sc.sid,sname,round(avg(score),2) as avg_score
from sc,student
where sc.sid = student.sid
group by sc.sid,sname
having avg_score>60

+------+--------+-----------+
| sid  | sname  | avg_score |
+------+--------+-----------+
| 01   | 赵雷   |     89.67 |
| 02   | 钱电   |     70.00 |
| 03   | 孙⻛   |     80.00 |
| 05   | 周梅   |     81.50 |
| 07   | 郑⽵   |     93.50 |
+------+--------+-----------+

6.查询在 SC 表存在成绩的学⽣信息

select distinct stu.*
from student as stu
join sc on sc.sid=stu.sid;

7.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 stu.sid,stu.sname,count(sc.cid) as num,sum(sc.score) as total_score
from student as stu
left join sc on sc.sid=stu.sid
group by stu.sid,stu.sname;

+------+--------+-----+-------------+
| sid  | sname  | num | total_score |
+------+--------+-----+-------------+
| 01   | 赵雷   |   3 |       269.0 |
| 02   | 钱电   |   3 |       210.0 |
| 03   | 孙⻛   |   3 |       240.0 |
| 04   | 李云   |   3 |       100.0 |
| 05   | 周梅   |   2 |       163.0 |
| 06   | 吴兰   |   2 |        65.0 |
| 07   | 郑⽵   |   2 |       187.0 |
| 09   | 张三   |   0 |        NULL |
| 10   | 李四   |   0 |        NULL |
| 11   | 李四   |   0 |        NULL |
| 12   | 赵六   |   0 |        NULL |
| 13   | 孙七   |   0 |        NULL |
+------+--------+-----+-------------+

8.查询「李」姓⽼师的数量

select count(*) as '姓李的老师的数量'
from teacher
where tname like '李%'

+--------------------------+
| 姓李的老师的数量         |
+--------------------------+
|                        1 |
+--------------------------+

9.查询学过「张三」⽼师授课的同学的信息

select student.* , teacher.tname
from student
join sc on student.sid=sc.sid
join course on course.cid=sc.cid
join teacher on teacher.tid=course.tid
where teacher.tname='张三'

+------+--------+---------------------+------+--------+
| SId  | Sname  | Sage                | Ssex | tname  |
+------+--------+---------------------+------+--------+
| 01   | 赵雷   | 1990-01-01 00:00:00 | 男   | 张三   |
| 02   | 钱电   | 1990-12-21 00:00:00 | 男   | 张三   |
| 03   | 孙⻛   | 1990-12-20 00:00:00 | 男   | 张三   |
| 04   | 李云   | 1990-12-06 00:00:00 | 男   | 张三   |
| 05   | 周梅   | 1991-12-01 00:00:00 | ⼥   | 张三   |
| 07   | 郑⽵   | 1989-01-01 00:00:00 | ⼥   | 张三   |
+------+--------+---------------------+------+--------+

10.查询没有学全所有课程的同学的信息

select stu.* ,count(sc.cid) as counts
from student stu
join sc on sc.sid=stu.sid
group by stu.sid,stu.sname,stu.Ssex,stu.Sage
having counts<(select count(*) from course)

+------+--------+---------------------+------+--------+
| SId  | Sname  | Sage                | Ssex | counts |
+------+--------+---------------------+------+--------+
| 05   | 周梅   | 1991-12-01 00:00:00 | ⼥   |      2 |
| 06   | 吴兰   | 1992-01-01 00:00:00 | ⼥   |      2 |
| 07   | 郑⽵   | 1989-01-01 00:00:00 | ⼥   |      2 |
+------+--------+---------------------+------+--------+

11.查询⾄少有⼀⻔课与学号为" 01 "的同学所学相同的同学的信息

思路：先查询01同学选过的课，然后判断是否 in 这查询结果中，我们会得到id重复的信息，这里还需要去重 distinct 

select distinct student.*
from student
join sc on sc.sid=student.sid
where sc.cid in (select sc.cid from sc where sc.sid='01') and sc.sid<>'01'

+------+--------+---------------------+------+
| SId  | Sname  | Sage                | Ssex |
+------+--------+---------------------+------+
| 02   | 钱电   | 1990-12-21 00:00:00 | 男   |
| 03   | 孙⻛   | 1990-12-20 00:00:00 | 男   |
| 04   | 李云   | 1990-12-06 00:00:00 | 男   |
| 05   | 周梅   | 1991-12-01 00:00:00 | ⼥   |
| 06   | 吴兰   | 1992-01-01 00:00:00 | ⼥   |
| 07   | 郑⽵   | 1989-01-01 00:00:00 | ⼥   |
+------+--------+---------------------+------+

12.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

思路：先查询01同学选过的课，然后查询同学的课程等于01同学选过课程的匹配数量，在比较与01同学的课程数是否一致

关键点是两个成绩表分别取别名，一个表查是01同学的另一个表查不是01同学的，还有 group by 匹配到的课程数

select s2.sid,student.sname,student.Sage,student.Ssex
from sc s1
join sc s2 
on s1.cid =s2.cid and s1.sid='01' and s2.sid!='01'
join student on student.sid = s2.sid
group by s2.sid,student.sname,student.Sage,student.Ssex
having count(s2.sid) = (select count(*) from sc where sid='01');

+------+--------+---------------------+------+
| sid  | sname  | Sage                | Ssex |
+------+--------+---------------------+------+
| 02   | 钱电   | 1990-12-21 00:00:00 | 男   |
| 03   | 孙⻛   | 1990-12-20 00:00:00 | 男   |
| 04   | 李云   | 1990-12-06 00:00:00 | 男   |
+------+--------+---------------------+------+

13.查询没学过"张三"⽼师讲授的任⼀⻔课程的学⽣姓名

这是之前的查询学习过张三老师课程的SQL，这里我们取反

select student.* , teacher.tname
from student
join sc on student.sid=sc.sid
join course on course.cid=sc.cid
join teacher on teacher.tid=course.tid
where teacher.tname='张三'

+------+--------+---------------------+------+--------+
| SId  | Sname  | Sage                | Ssex | tname  |
+------+--------+---------------------+------+--------+
| 01   | 赵雷   | 1990-01-01 00:00:00 | 男   | 张三   |
| 02   | 钱电   | 1990-12-21 00:00:00 | 男   | 张三   |
| 03   | 孙⻛   | 1990-12-20 00:00:00 | 男   | 张三   |
| 04   | 李云   | 1990-12-06 00:00:00 | 男   | 张三   |
| 05   | 周梅   | 1991-12-01 00:00:00 | ⼥   | 张三   |
| 07   | 郑⽵   | 1989-01-01 00:00:00 | ⼥   | 张三   |
+------+--------+---------------------+------+--------+

查询不在这个列表即可

select student.* 
from student
where student.sid not in(
select student.sid
from student
join sc on student.sid=sc.sid
join course on course.cid=sc.cid
join teacher on teacher.tid=course.tid
where teacher.tname='张三'
)

+------+--------+---------------------+------+ 
| SId  | Sname  | Sage                | Ssex | 
+------+--------+---------------------+------+ 
| 06   | 吴兰   | 1992-01-01 00:00:00 | ⼥   |    
| 09   | 张三   | 2017-12-20 00:00:00 | ⼥   |    
| 10   | 李四   | 2017-12-25 00:00:00 | ⼥   |    
| 11   | 李四   | 2012-06-06 00:00:00 | ⼥   |    
| 12   | 赵六   | 2013-06-13 00:00:00 | ⼥   |    
| 13   | 孙七   | 2014-06-01 00:00:00 | ⼥   |    
+------+--------+---------------------+------+

14.查询两⻔及其以上不及格课程的同学的学号，姓名及其平均成绩

先统计不及格的分数，再进行分组，再计算不及格趁机的数量是否大于等于2

select stu.sid,stu.sname,round(avg(sc.score),2) as avg_score
from student as stu
join sc on stu.sid=sc.sid
where sc.score<60
group by stu.sid,stu.sname 
having count(sc.cid)>=2;

+------+--------+-----------+
| sid  | sname  | avg_score |
+------+--------+-----------+
| 04   | 李云   |     33.33 |
| 06   | 吴兰   |     32.50 |
+------+--------+-----------+

15.检索" 01 "课程分数⼩于 60，按分数降序排列的学⽣信息

select student.sname,student.Sage,student.Ssex,sc.score
from student
join sc on student.sid=sc.sid and sc.cid='01' and sc.score<60
group by sc.score,student.sname,student.Sage,student.Ssex desc

+--------+---------------------+------+-------+
| sname  | Sage                | Ssex | score |
+--------+---------------------+------+-------+
| 吴兰   | 1992-01-01 00:00:00 | ⼥   |  31.0 |
| 李云   | 1990-12-06 00:00:00 | 男   |  50.0 |
+--------+---------------------+------+-------+

16.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

思路：先查询所有学生的平均成绩，再和sc表管理，显示所有课程的成绩

select sid,avg(score) avg_score 
from sc
group by sid

+------+-----------+ 
| sid  | avg_score | 
+------+-----------+ 
| 01   |  89.66667 | 
| 02   |  70.00000 | 
| 03   |  80.00000 | 
| 04   |  33.33333 | 
| 05   |  81.50000 | 
| 06   |  32.50000 | 
| 07   |  93.50000 | 
+------+-----------+ 

关联

select sc.*,s2.avg_score
from sc 
join (
	select sid,avg(score) avg_score 
	from sc
	group by sid
)as s2
on sc.sid=s2.sid
order by s2.avg_score desc,sc.sid;

+------+------+-------+-----------+
| SId  | CId  | score | avg_score |
+------+------+-------+-----------+
| 07   | 02   |  89.0 |  93.50000 |
| 07   | 03   |  98.0 |  93.50000 |
| 01   | 02   |  90.0 |  89.66667 |
| 01   | 03   |  99.0 |  89.66667 |
| 01   | 01   |  80.0 |  89.66667 |
| 05   | 01   |  76.0 |  81.50000 |
| 05   | 02   |  87.0 |  81.50000 |
| 03   | 01   |  80.0 |  80.00000 |
| 03   | 02   |  80.0 |  80.00000 |
| 03   | 03   |  80.0 |  80.00000 |
| 02   | 02   |  60.0 |  70.00000 |
| 02   | 03   |  80.0 |  70.00000 |
| 02   | 01   |  70.0 |  70.00000 |
| 04   | 01   |  50.0 |  33.33333 |
| 04   | 02   |  30.0 |  33.33333 |
| 04   | 03   |  20.0 |  33.33333 |
| 06   | 03   |  34.0 |  32.50000 |
| 06   | 01   |  31.0 |  32.50000 |
+------+------+-------+-----------+

我们看到每一次都打印了平均成绩，这里不能去重了，去重我们就看不到所有的课程成绩了

思路：分别取三个别名，把三科的成绩都关联起来

 select stu.sid,stu.sname ,a.score,b.score,c.score,round(avg(d.score),2) as avg_score
 from student as stu 
 left join sc as a on stu.sid=a.sid and a.cid='01' 
 left join sc as b on stu.sid=b.sid and b.cid='02'
 left join sc as c on stu.sid=c.sid and c.cid='03'
 left join sc as d on stu.sid=d.sid
 group by stu.sid,stu.sname,a.score,b.score,c.score
 order by avg_score desc

+------+--------+-------+-------+-------+-----------+
| sid  | sname  | score | score | score | avg_score |
+------+--------+-------+-------+-------+-----------+
| 07   | 郑⽵   |  NULL |  89.0 |  98.0 |     93.50 |
| 01   | 赵雷   |  80.0 |  90.0 |  99.0 |     89.67 |
| 05   | 周梅   |  76.0 |  87.0 |  NULL |     81.50 |
| 03   | 孙⻛   |  80.0 |  80.0 |  80.0 |     80.00 |
| 02   | 钱电   |  70.0 |  60.0 |  80.0 |     70.00 |
| 04   | 李云   |  50.0 |  30.0 |  20.0 |     33.33 |
| 06   | 吴兰   |  31.0 |  NULL |  34.0 |     32.50 |
| 09   | 张三   |  NULL |  NULL |  NULL |      NULL |
| 11   | 李四   |  NULL |  NULL |  NULL |      NULL |
| 13   | 孙七   |  NULL |  NULL |  NULL |      NULL |
| 10   | 李四   |  NULL |  NULL |  NULL |      NULL |
| 12   | 赵六   |  NULL |  NULL |  NULL |      NULL |
+------+--------+-------+-------+-------+-----------+

用了4个 left join 但是结果比上一种更加的好看一点

17.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 
#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


如果单独求最高分最低分，平均值，那这道题非常简单了。
select max(score),min(score),avg(score)
from sc
+------------+------------+------------+
| max(score) | min(score) | avg(score) |
+------------+------------+------------+
|       99.0 |       20.0 |   68.55556 |
+------------+------------+------------+

但是我发现题目好像没有显示完，往后拉，先一步一步来。

select sc.cid,c.cname,
max(sc.score) as '最高分',
min(sc.score) as '最低分',
avg(sc.score) as '平均分',
count(sc.cid) as '选课人数'
from sc
join course as c on sc.cid=c.cid
group by sc.cid,c.cname
order by '选修人数' desc,sc.cid;

+------+--------+-----------+-----------+-----------+--------------+
| cid  | cname  | 最高分    | 最低分    | 平均分    | 选课人数     |
+------+--------+-----------+-----------+-----------+--------------+
| 01   | 语⽂   |      80.0 |      31.0 |  64.50000 |            6 |
| 02   | 数学   |      90.0 |      30.0 |  72.66667 |            6 |
| 03   | 英语   |      99.0 |      20.0 |  68.50000 |            6 |
+------+--------+-----------+-----------+-----------+--------------+

对于及格率，优秀率的计算，这里需要使用我们的新语法 case then '条件' then '满足条件的结果' else '不满足条件的结果' end

case when sc.score>=60 then 1 else 0 end 
相当于java中的
if(sc.score>=60){
	return 1;
}
else{
	reuturn 0;
}

select sc.cid,c.cname,
max(sc.score) as '最高分',
min(sc.score) as '最低分',
round(avg(sc.score),2) as '平均分',
count(sc.cid) as '选课人数',
round(sum(case when sc.score>=60 then 1 else 0 end)/count(sc.cid),2) as '及格率', 
round(sum(case when sc.score>=70 and sc.score<80 then 1 else 0 end)/count(sc.cid),2) as '中等率',
round(sum(case when sc.score>=80 and sc.score<90 then 1 else 0 end)/count(sc.cid),2) as '优良率',
round(sum(case when sc.score>=90 then 1 else 0 end)/count(sc.cid),2) as '优秀率'
from sc
join course as c on sc.cid=c.cid
group by sc.cid,c.cname
order by '选修人数' desc,sc.cid;

+------+--------+-----------+-----------+-----------+--------------+-----------+-----------+-----------+-----------+
| cid  | cname  | 最高分    | 最低分    | 平均分    | 选课人数     | 及格率    | 中等率    | 优良率    | 优秀率    |
+------+--------+-----------+-----------+-----------+--------------+-----------+-----------+-----------+-----------+
| 01   | 语⽂   |      80.0 |      31.0 |     64.50 |            6 |      0.67 |      0.33 |      0.33 |      0.00 |
| 02   | 数学   |      90.0 |      30.0 |     72.67 |            6 |      0.83 |      0.00 |      0.50 |      0.17 |
| 03   | 英语   |      99.0 |      20.0 |     68.50 |            6 |      0.67 |      0.00 |      0.33 |      0.33 |
+------+--------+-----------+-----------+-----------+--------------+-----------+-----------+-----------+-----------+

18.按各科平均成绩进⾏排序，并显示排名， Score 重复时保留名次空缺

先获取各科的平均成绩并排序

select cid,round(avg(sc.score),2) as avg_score 
from sc 
group by cid
order by avg_score 
+------+-----------+
| cid  | avg_score |
+------+-----------+
| 01   |     64.50 |
| 03   |     68.50 |
| 02   |     72.67 |
+------+-----------+ 

那么如何获得排名呢？

select s1.*,s2.* 
from 
( 
select cid,round(avg(sc.score),2) as avg_score 
from sc 
group by cid
)as s1 
join ( 
select cid,round(avg(sc.score),2) as avg_score 
from sc 
group by cid
)as s2 on s1.avg_score>=s2.avg_score;

自连接

+------+-----------+------+-----------+
| cid  | avg_score | cid  | avg_score |
+------+-----------+------+-----------+
| 01   |     64.50 | 01   |     64.50 |
| 02   |     72.67 | 01   |     64.50 |
| 03   |     68.50 | 01   |     64.50 |
| 02   |     72.67 | 02   |     72.67 |
| 02   |     72.67 | 03   |     68.50 |
| 03   |     68.50 | 03   |     68.50 |
+------+-----------+------+-----------+



按照s2进行分组，然后统计s1的平均分出现的次数，
为什么要这么搞呢？我们的关联条件 s1.avg_score>=s2.avg_score 中
最大的数字只能找到自己满足条件 值为 1
倒数第二大的数组可以找到自己和最大的数 值为 2
以此类推，得到的结果就是排名

select s2.cid,s2.avg_score,count(distinct s1.avg_score) as rank
from
(
select cid,round(avg(sc.score),2) as avg_score 
from sc 
group by cid
)as s1
join
(
select cid,round(avg(sc.score),2) as avg_score 
from sc 
group by cid
)as s2 
on s1.avg_score>=s2.avg_score
group by s2.cid,s2.avg_score
order by rank;

+------+-----------+------+
| cid  | avg_score | rank |
+------+-----------+------+
| 02   |     72.67 |    1 |
| 03   |     68.50 |    2 |
| 01   |     64.50 |    3 |
+------+-----------+------+

题目要求： Score 重复时保留名次空缺
这里我们三个课程成绩不一样，我们临时修改一下看一下结果

begin;
update sc set score =104.0 where sid=01 and cid=01;
select s2.cid,s2.avg_score,count(distinct s1.avg_score) as rank
from
(
select cid,round(avg(sc.score),2) as avg_score 
from sc 
group by cid
)as s1
join
(
select cid,round(avg(sc.score),2) as avg_score 
from sc 
group by cid
)as s2 
on s1.avg_score>=s2.avg_score
group by s2.cid,s2.avg_score
order by rank;	
rollback;

Score 重复时不保留名次空缺
+------+-----------+------+
| cid  | avg_score | rank |
+------+-----------+------+
| 02   |     72.67 |    1 |
| 03   |     68.50 |    2 |
| 01   |     68.50 |    2 |
+------+-----------+------+

begin;
update sc set sc.score=50 where sc.sid=06;

select s1.sid,s1.sname,s1.total_score,count(s1.total_score) as rank
from (
select student.sid,student.sname,sum(sc.score) as total_score
from student
join sc on sc.sid=student.sid
group by student.sid,student.sname
order by total_score
) as s1
join (
select student.sid,student.sname,sum(sc.score) as total_score
from student
join sc on sc.sid=student.sid
group by student.sid,student.sname
order by total_score
) as s2
on s1.total_score<=s2.total_score
group by s1.sid,s1.sname,s1.total_score
order by rank;
rollback;

//Help
+------+--------+-------------+------+  
| sid  | sname  | total_score | rank |  
+------+--------+-------------+------+  
| 01   | 赵雷   |       269.0 |    1 |    
| 03   | 孙⻛   |       240.0 |    2 |    
| 02   | 钱电   |       210.0 |    3 |    
| 07   | 郑⽵   |       187.0 |    4 |    
| 05   | 周梅   |       163.0 |    5 |    
| 06   | 吴兰   |       100.0 |    7 |    
| 04   | 李云   |       100.0 |    7 |    
+------+--------+-------------+------+

19.按各科平均成绩进⾏排序，并显示排名， Score 重复时不保留名次空缺


我们发现第一个数据排名是1，第二是2，那么我们有没有什么简单的方法呢？

由于我们表中没有重复的数据，这里我们临时改一下数据，定义变量：@i

begin;
#临时修改数据
update sc set score =104.0 where sid=01 and cid=01;

select b.cid,b.avg_score,@i:=@i+1 as rank
from 
(select @i:=0) as a,
(
	select cid,round(avg(score),2) as avg_score
	from sc 
	group by cid
	order by avg_score desc
)
as b;
rollback;


Score 重复时不保留名次空缺
+------+-----------+------+ 
| cid  | avg_score | rank | 
+------+-----------+------+ 
| 02   |     72.67 |    1 | 
| 01   |     68.50 |    2 | 
| 03   |     68.50 |    3 | 
+------+-----------+------+

20.查询学⽣的总成绩，并进⾏排名，总分重复时保留名次空缺

由于我们表中没有重复的总分，这里我们临时改一下数据

begin;
update sc set sc.score=50 where sc.sid=06;
select student.sid,student.sname,sum(sc.score) as total_score
from student
join sc on sc.sid=student.sid
group by student.sid,student.sname
order by total_score;
rollback;

+------+--------+-------------+
| sid  | sname  | total_score |
+------+--------+-------------+
| 06   | 吴兰   |       100.0 |
| 04   | 李云   |       100.0 |
| 05   | 周梅   |       163.0 |
| 07   | 郑⽵   |       187.0 |
| 02   | 钱电   |       210.0 |
| 03   | 孙⻛   |       240.0 |
| 01   | 赵雷   |       269.0 |
+------+--------+-------------+

题目要求：总分重复时保留名次空缺

begin;
update sc set sc.score=50 where sc.sid=06;

select s1.sid,s1.sname,s1.total_score,count(distinct s1.total_score) as rank,
from (
select student.sid,student.sname,sum(sc.score) as total_score
from student
join sc on sc.sid=student.sid
group by student.sid,student.sname
order by total_score;
) as s1
join (
select student.sid,student.sname,sum(sc.score) as total_score
from student
join sc on sc.sid=student.sid
group by student.sid,student.sname
order by total_score;	
) as s2
on s1.total_score>=s2.total_score;
group by s1.sid,s1.sname,s1.total_score
order by rank;
rollback;

21.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

使用临时变量来实现

begin;
update sc set sc.score=50 where sc.sid=06;

select b.sid,b.total_score,@i:=@i+1 as rank
from 
(select @i:=0) as a,
(
	select sid,sum(score) as total_score
	from sc 
	group by sid
	order by total_score desc
)
as b;
rollback;

不保留名次空缺

+------+-------------+------+
| sid  | total_score | rank |
+------+-------------+------+
| 01   |       269.0 |    1 |
| 03   |       240.0 |    2 |
| 02   |       210.0 |    3 |
| 07   |       187.0 |    4 |
| 05   |       163.0 |    5 |
| 04   |       100.0 |    6 |
| 06   |       100.0 |    7 |
+------+-------------+------+

22.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 sc.cid,c.cname,
max(sc.score) as '最高分',
min(sc.score) as '最低分',
round(avg(sc.score),2) as '平均分',
count(sc.cid) as '选课人数',
round(sum(case when sc.score>=85 then 1 else 0 end)/count(sc.cid),2) as '[100-85]', 
round(sum(case when sc.score>=70 and sc.score<85 then 1 else 0 end)/count(sc.cid),2) as '[85-70]',
round(sum(case when sc.score>=60 and sc.score<70 then 1 else 0 end)/count(sc.cid),2) as '[70-60]',
round(sum(case when sc.score>=0  and sc.score<60 then 1 else 0 end)/count(sc.cid),2) as '[60-0]'
from sc
join course as c on sc.cid=c.cid
group by sc.cid,c.cname
order by '选修人数' desc,sc.cid;

+------+--------+-----------+-----------+-----------+--------------+----------+---------+---------+--------+
| cid  | cname  | 最高分    | 最低分    | 平均分    | 选课人数     | [100-85] | [85-70] | [70-60] | [60-0] |
+------+--------+-----------+-----------+-----------+--------------+----------+---------+---------+--------+
| 01   | 语⽂   |      80.0 |      31.0 |     64.50 |            6 |     0.00 |    0.67 |    0.00 |   0.33 |
| 02   | 数学   |      90.0 |      30.0 |     72.67 |            6 |     0.50 |    0.17 |    0.17 |   0.17 |
| 03   | 英语   |      99.0 |      20.0 |     68.50 |            6 |     0.33 |    0.33 |    0.00 |   0.33 |
+------+--------+-----------+-----------+-----------+--------------+----------+---------+---------+--------+

23.查询各科成绩前三名的记录

union 加 limit 来完成

select s1.* from(select sid,cid,score 
	from sc where cid =01
	GROUP BY sid,cid,score 
	order by score desc limit 3
) as s1
union
select s2.* from (select sid,cid,score
	from sc where cid =02 
	GROUP BY sid,cid,score 
	order by score desc limit 3 
) as s2
union
select s3.* from (select sid,cid,score 
	from sc where cid =03 
	GROUP BY sid,cid,score 
	order by score desc limit 3 
) as s3;

+------+------+-------+
| sid  | cid  | score |
+------+------+-------+
| 03   | 01   |  80.0 |
| 01   | 01   |  80.0 |
| 05   | 01   |  76.0 |
| 01   | 02   |  90.0 |
| 07   | 02   |  89.0 |
| 05   | 02   |  87.0 |
| 01   | 03   |  99.0 |
| 07   | 03   |  98.0 |
| 02   | 03   |  80.0 |
+------+------+-------+

24.查询每⻔课程被选修的学⽣数

select sc.cid,count(sc.cid)
from sc
group by sc.cid;

+------+---------------+
| cid  | count(sc.cid) |
+------+---------------+
| 01   |             6 |
| 02   |             6 |
| 03   |             6 |
+------+---------------+

25.查询出只选修两⻔课程的学⽣学号和姓名

select stu.sid,stu.sname
from student as stu
join sc on stu.sid=sc.sid
group by stu.sid,stu.sname
having count(sc.cid)=2

+------+--------+
| sid  | sname  |
+------+--------+
| 05   | 周梅   |
| 06   | 吴兰   |
| 07   | 郑⽵   |
+------+--------+

26.查询男⽣、⼥⽣⼈数

select count(*) from student where Ssex='男'; 
select count(*) from student where Ssex='女'; 

+----------+
| count(*) |
+----------+
|        4 |
+----------+

+----------+
| count(*) |
+----------+
|        8 |
+----------+

如果要一起显示，可以这么写
select ssex,count(ssex) as 数量 from student GROUP BY ssex;

+------+--------+
| ssex | 数量   |
+------+--------+
| ⼥   |      8 |
| 男   |      4 |
+------+--------+

27.查询名字中含有「⻛」字的学⽣信息

select * from student where sname like '%风%';

+------+--------+---------------------+------+
| SId  | Sname  | Sage                | Ssex |
+------+--------+---------------------+------+
| 03   | 孙⻛   | 1990-12-20 00:00:00 | 男   |
+------+--------+---------------------+------+

28.查询同名同性学⽣名单，并统计同名⼈数

select count(*) as '同名同姓人数'
from student as s1
join student as s2 on s1.sname=s2.sname and s1.Ssex=s2.Ssex and s1.sid!=s2.sid

+--------------------+
| 同名同姓人数       |
+--------------------+
|                  2 |
+--------------------+

29.查询 1990 年出⽣的学⽣名单

+------+--------+---------------------+------+
| SId  | Sname  | Sage                | Ssex |
+------+--------+---------------------+------+
| 01   | 赵雷   | 1990-01-01 00:00:00 | 男   |
| 02   | 钱电   | 1990-12-21 00:00:00 | 男   |
| 03   | 孙⻛   | 1990-12-20 00:00:00 | 男   |
| 04   | 李云   | 1990-12-06 00:00:00 | 男   |
+------+--------+---------------------+------+

30.查询每⻔课程的平均成绩，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 sc.cid,avg(sc.score) as avg_score
from sc
group by sc.cid
order by avg_score desc,sc.cid;

+------+-----------+
| cid  | avg_score |
+------+-----------+
| 02   |  72.66667 |
| 03   |  68.50000 |
| 01   |  64.50000 |
+------+-----------+

31.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 stu.sid,stu.sname,avg(sc.score) as '平均成绩'
from student as stu
join sc on sc.sid=stu.sid
group by stu.sid,stu.sname
having avg(sc.score)>=85;

+------+--------+--------------+
| sid  | sname  | 平均成绩     |
+------+--------+--------------+
| 01   | 赵雷   |     89.66667 |
| 07   | 郑⽵   |     93.50000 |
+------+--------+--------------+

32.查询课程名称为「数学」，且分数低于 60 的学⽣姓名和分数

select stu.sname,sc.score,c.cname
from sc
join student as stu on stu.sid=sc.sid
join course as c on c.cid=sc.cid and c.cname='数学'
where sc.score<60;

33.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）

存在学⽣没成绩，没选课的情况,我们应该把学生表作为主表，一门课都没选的学生则用NULL表示

select stu.sid,c.cname,sc.score
from student as stu
left join sc on stu.sid=sc.sid
left join course as c on sc.cid=c.cid
order by stu.sid;

34.查询任何⼀⻔课程成绩在 70 分以上的姓名、课程名称和分数

select stu.sname,c.cname,sc.score
from sc
join student as stu on stu.sid=sc.sid
join course as c on c.cid=sc.cid
where sc.score>70;

+--------+--------+-------+
| sname  | cname  | score |
+--------+--------+-------+
| 赵雷   | 语⽂   |  80.0 |
| 赵雷   | 数学   |  90.0 |
| 赵雷   | 英语   |  99.0 |
| 钱电   | 英语   |  80.0 |
| 孙⻛   | 语⽂   |  80.0 |
| 孙⻛   | 数学   |  80.0 |
| 孙⻛   | 英语   |  80.0 |
| 周梅   | 语⽂   |  76.0 |
| 周梅   | 数学   |  87.0 |
| 郑⽵   | 数学   |  89.0 |
| 郑⽵   | 英语   |  98.0 |
+--------+--------+-------+

35.查询不及格的课程

select cid,count(score) as '不及格'
from sc
where score<60
group by cid;

+------+-----------+
| cid  | 不及格    |
+------+-----------+
| 01   |         2 |
| 02   |         1 |
| 03   |         2 |
+------+-----------+

36.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 stu.sid,stu.sname,sc.score as '分数'
from student as stu
join sc on sc.sid=stu.sid and sc.cid='01' and sc.score>=80;

+------+--------+-------+
| sid  | sname  | 分数 |
+------+--------+-------+
| 01   | 赵雷   |  80.0 |
| 03   | 孙⻛   |  80.0 |
+------+--------+-------+

37.求每⻔课程的学⽣⼈数

select cid,count(cid) as '学生人数'
from sc
group by cid;

+------+--------------+
| cid  | 学生人数     |
+------+--------------+
| 01   |            6 |
| 02   |            6 |
| 03   |            6 |
+------+--------------+

38.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 student.*,sc.score
from student
join sc on sc.sid=student.sid and sc.cid='02' and sc.score = (
select max(score)
from student as stu
join sc on sc.sid=stu.sid
join course as c on c.cid=sc.cid
join teacher as t on t.tid=c.tid
where t.tname='张三'
);

+------+--------+---------------------+------+-------+
| SId  | Sname  | Sage                | Ssex | score |
+------+--------+---------------------+------+-------+
| 01   | 赵雷   | 1990-01-01 00:00:00 | 男   |  90.0 |
+------+--------+---------------------+------+-------+

39.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

由于我们表中没有重复的总分，这里我们临时改一下数据
begin;
update sc set score=90 where sid=04 and cid=02;

select student.*,sc.score
from student
join sc on sc.sid=student.sid and sc.cid='02' and sc.score = (
select max(score)
from student as stu
join sc on sc.sid=stu.sid
join course as c on c.cid=sc.cid
join teacher as t on t.tid=c.tid
where t.tname='张三'
);
rollback;

+------+--------+---------------------+------+-------+
| SId  | Sname  | Sage                | Ssex | score |
+------+--------+---------------------+------+-------+
| 01   | 赵雷   | 1990-01-01 00:00:00 | 男   |  90.0 |
| 04   | 李云   | 1990-12-06 00:00:00 | 男   |  90.0 |
+------+--------+---------------------+------+-------+

40.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 s1.sid,s1.cid, s2.score
from sc as s1
join sc as s2 on s1.sid=s2.sid and s1.cid!=s2.cid and s1.score=s2.score;

+------+------+-------+       
| sid  | cid  | score |       
+------+------+-------+       
| 03   | 02   |  80.0 |       
| 03   | 03   |  80.0 |       
| 03   | 01   |  80.0 |       
| 06   | 03   |  50.0 |       
| 06   | 01   |  50.0 |       
+------+------+-------+

41.查询每⻔课程成绩最好的前两名

这里用union连接，每次limit两个

select s1.* 
from(
	select sid,cid,score 
	from sc where cid =01 
	GROUP BY sid,cid,score 
	order by score desc limit 2
) as s1
union
select s2.* 
from(
	select sid,cid,score 
	from sc where cid =02 
	GROUP BY sid,cid,score 
	order by score desc limit 2
) as s2
union
select s3.* 
from(
	select sid,cid,score 
	from sc where cid =03 
	GROUP BY sid,cid,score 
	order by score desc limit 2
) as s3

+------+------+-------+
| sid  | cid  | score |
+------+------+-------+
| 03   | 01   |  80.0 |
| 01   | 01   |  80.0 |
| 01   | 02   |  90.0 |
| 07   | 02   |  89.0 |
| 01   | 03   |  99.0 |
| 07   | 03   |  98.0 |
+------+------+-------+

42.统计每⻔课程的学⽣选修⼈数（超过 5 ⼈的课程才统计）。

select cid,count(cid) as '学生人数'
from sc
group by cid
having count(cid)>5;

+------+--------------+
| cid  | 学生人数     |
+------+--------------+
| 01   |            6 |
| 02   |            6 |
| 03   |            6 |
+------+--------------+

43.检索⾄少选修两⻔课程的学⽣学号

select sc.sid,count(sc.sid) as '课程数'
from sc
group by sc.sid
having count(sc.sid)>=2;

+------+-----------+  
| sid  | 课程数    |     
+------+-----------+  
| 01   |         3 |  
| 02   |         3 |  
| 03   |         3 |  
| 04   |         3 |  
| 05   |         2 |  
| 06   |         2 |  
| 07   |         2 |  
+------+-----------+

44.查询选修了全部课程的学⽣信息

select sc.sid,count(sc.sid) as '课程数'
from sc
group by sc.sid
having count(sc.sid)=(select count(*) from course);

+------+-----------+
| sid  | 课程数    |
+------+-----------+
| 01   |         3 |
| 02   |         3 |
| 03   |         3 |
| 04   |         3 |
+------+-----------+

45.查询各学⽣的年龄，只按年份来算

思路：用year函数将now时间与学生时间转换然后相减

select sname,year(now())-year(sage)as 年龄 from student order by sage;

+--------+--------+ 
| sname  | 年龄   |   
+--------+--------+ 
| 郑⽵   |     31 |   
| 赵雷   |     30 |   
| 李云   |     30 |   
| 孙⻛   |     30 |   
| 钱电   |     30 |   
| 周梅   |     29 |   
| 吴兰   |     28 |   
| 李四   |      8 |   
| 赵六   |      7 |   
| 孙七   |      6 |   
| 张三   |      3 |   
| 李四   |      3 |   
+--------+--------+

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

使用 TIMESTAMPDIFF 函数，传入年，Sage和现在的时间，计算年份

select sid,sname,TIMESTAMPDIFF(YEAR,Sage,NOW()) as age from student;

+------+--------+------+ 
| sid  | sname  | age  | 
+------+--------+------+ 
| 01   | 赵雷   |   30 |   
| 02   | 钱电   |   29 |   
| 03   | 孙⻛   |   29 |   
| 04   | 李云   |   29 |   
| 05   | 周梅   |   28 |   
| 06   | 吴兰   |   28 |   
| 07   | 郑⽵   |   31 |   
| 09   | 张三   |    2 |   
| 10   | 李四   |    2 |   
| 11   | 李四   |    8 |   
| 12   | 赵六   |    7 |   
| 13   | 孙七   |    6 |   
+------+--------+------+

47.查询本周过⽣⽇的学⽣

思路：学习week的使用
没有数据，临时改一下

begin;
update student set sage=now() where sid='01';
SELECT * 
FROM student where
week(Sage)=week(now());
rollback;

+------+--------+---------------------+------+
| SId  | Sname  | Sage                | Ssex |
+------+--------+---------------------+------+
| 01   | 赵雷   | 2020-09-14 20:00:16 | 男   |
+------+--------+---------------------+------+

48、查询下周过生日的学生

日期加减法比较麻烦，这里只给代码了

SELECT * 
FROM student where
week(Sage)=week(now())+1;

49.查询本⽉过⽣⽇的学⽣

思路：学习month的使用
没有数据，临时改一下

begin;
update student set sage=now() where sid='01';
SELECT * 
FROM student where
month(Sage)=month(now());
rollback;

+------+--------+---------------------+------+
| SId  | Sname  | Sage                | Ssex |
+------+--------+---------------------+------+
| 01   | 赵雷   | 2020-09-14 20:01:47 | 男   |
+------+--------+---------------------+------+

50.查询下⽉过⽣⽇的学⽣

日期加减法比较麻烦，这里只给代码了

SELECT * 
FROM student
where month(Sage)=month(now())+1;

本文地址：https://blog.csdn.net/weixin_44141495/article/details/108588964

MySQl经典50题，全部都会做就是大佬了！

MySQL经典50题，全部都会做就是大佬了！

文章目录

创建表结构

建表语句

添加数据

表关系

题目

参考答案

1.查询" 01 “课程⽐” 02 "课程成绩⾼的学⽣的信息及课程分数

2.查询同时存在" 01 “课程和” 02 "课程的情况

3.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 )

4.查询不存在" 01 “课程但存在” 02 "课程的情况

5.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

6.查询在 SC 表存在成绩的学⽣信息

7.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 )

8.查询「李」姓⽼师的数量

9.查询学过「张三」⽼师授课的同学的信息

10.查询没有学全所有课程的同学的信息

11.查询⾄少有⼀⻔课与学号为" 01 "的同学所学相同的同学的信息

12.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

13.查询没学过"张三"⽼师讲授的任⼀⻔课程的学⽣姓名

14.查询两⻔及其以上不及格课程的同学的学号，姓名及其平均成绩

15.检索" 01 "课程分数⼩于 60，按分数降序排列的学⽣信息

16.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

17.查询各科成绩最⾼分、最低分和平均分。以如下形式显示：

18.按各科平均成绩进⾏排序，并显示排名， Score 重复时保留名次空缺

19.按各科平均成绩进⾏排序，并显示排名， Score 重复时不保留名次空缺

20.查询学⽣的总成绩，并进⾏排名，总分重复时保留名次空缺

21.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

22.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

23.查询各科成绩前三名的记录

24.查询每⻔课程被选修的学⽣数

25.查询出只选修两⻔课程的学⽣学号和姓名

26.查询男⽣、⼥⽣⼈数

27.查询名字中含有「⻛」字的学⽣信息

28.查询同名同性学⽣名单，并统计同名⼈数

29.查询 1990 年出⽣的学⽣名单

30.查询每⻔课程的平均成绩，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

31.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

32.查询课程名称为「数学」，且分数低于 60 的学⽣姓名和分数

33.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）

34.查询任何⼀⻔课程成绩在 70 分以上的姓名、课程名称和分数

35.查询不及格的课程

36.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

37.求每⻔课程的学⽣⼈数

38.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

39.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

40.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

41.查询每⻔课程成绩最好的前两名

42.统计每⻔课程的学⽣选修⼈数（超过 5 ⼈的课程才统计）。

43.检索⾄少选修两⻔课程的学⽣学号

44.查询选修了全部课程的学⽣信息

45.查询各学⽣的年龄，只按年份来算

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

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