MySQL 练习题目 二刷 - 2019-11-4 5:55 am

建表的过程

create table student(
sid int not null primary key,
sname varchar(20) not null,
sborn date,
ssex varchar(20) not null);

create table course(
cid int not null primary key,
cname varchar(20) not null,
tid int not null);

create table teacher(
tid int not null primary key,
tname varchar(20));

create table sc(
sid int not null,
cid int not null,
score int not null,
primary key( sid, cid)   );

插入数据到student表

insert into Student values(1 , '赵雷' , '1990-01-01' , '男');
insert into Student values(2 , '钱电' , '1990-12-21' , '男');
insert into Student values('' , '孙风' , '1990-12-20' , '男');
insert into Student values('' , '李云' , '1990-12-06' , '男');
insert into Student values('' , '周梅' , '1991-12-01' , '女');
insert into Student values('' , '吴兰' , '1992-01-01' , '女');
insert into Student values('' , '郑竹' , '1989-01-01' , '女');
insert into Student values('' , '张三' , '2017-12-20' , '女');
insert into Student values('' , '李四' , '2017-12-25' , '女');
insert into Student values('' , '李四' , '2012-06-06' , '女');
insert into Student values('' , '赵六' , '2013-06-13' , '女');
insert into Student values('' , '孙七' , '2014-06-01' , '女');

insert into Course values('' , '语文' , ''),('' , '数学' , ''),('' , '英语' , ''),(4,'物理',4);

insert into Teacher values('' , '张三'),('' , '李四'),('' , '王五'),(4,'孙杨');

insert into SC values('' , '' , 80),
('' , '' , 90),
('' , '' , 99),
(1,4,46),
('' , '' , 70),
('' , '' , 60),
('' , '' , 80),
(2,4,76),
('' , '' , 80),
('' , '' , 80),
('' , '' , 80),
('' , '' , 50),
('' , '' , 30),
('' , '' , 20),
(4,4,87),
('' , '' , 76),
('' , '' , 87),
('' , '' , 31),
('' , '' , 34),
(6,4,93),
('' , '' , 89),
('' , '' , 98);

解题过程：

0、按平均成绩降序查询所有学生的课程成绩，按如下形式显示：学生姓名  课程名  课程数   所有成绩平均分

select  student.sname,
max(case when cname='chinese' then  sc.score else  NULL end)  as 'chinese', #已经按学号分组了，所以每组就是每个同学的所有课程成绩
max(case when cname='math' then  sc.score else NULL end ) as 'math',
max(case when cname='english' then  sc.score else NULL end) as 'english',
max(case when cname='PE' then  sc.score else NULL end) as 'pe',
count(sc.cid) as '选课数',
avg(score) as '平均成绩'

from  sc join student on sc.sid=student.sid  join course on sc.cid=course.cid

group by sc.sid ,student.sname
order by avg(score) desc;

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 student.*,a.score

from  (sc a  join sc b  on  a.sid=b.sid
                        and a.cid=1
                        and b.cid=2
                        and a.score>b.score)
      join student on a.sid=student.sid;

2、查询同时选修" 01 "课程和" 02 "课程的学生情况

select student.*

from (sc  a join sc b on a.sid=b.sid and a.cid=1 and b.cid=2) join student on a.sid=student.sid;

这里应该是从笛卡尔积形式的连接中查询需要的值，但是使用Join连接时，只设置了a.sid=b.sid，没有写a.cid=b.cid，所以也能达到笛卡尔积形式的连接

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 )

#思路:题目的意思是选修1号课程的学生必须全部列出来，同时查看一下选修了1号课程的学生中有哪些选修了2号课程

select * from

(select *  from sc  where cid=1 ) a  left join  (select * from sc where cid=2) b on a.sid=b.sid ;

4、查询不存在" 01 "课程但存在" 02 "课程的情况

select a.sid  from 

(select sid from sc where cid=2 ) a left join (select sid from sc where cid=1) b on a.sid=b.sid
where b.sid is null;

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,student.sname, avg(sc.score) as '平均成绩'
from sc left join student on sc.sid=student.sid
group by sc.sid
having avg(sc.score)>=60;

6、查询在 SC 表存在成绩的学生信息

select distinct student.*
from sc left join student on sc.sid=student.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 )

#没成绩的显示null，说明有left join 或 right join，总有一个表内容要全部显示
select student.sid, student.sname, count(sc.cid) as '选课总数', sum(sc.score) as '总成绩'
from student left join sc on student.sid=sc.sid
group by sc.sid
order by student.sid;  #为了更直观，对sid排了序

8、查有成绩的学生信息

select distinct student.*
from student join sc on student.sid=sc.sid
where sc.score is not null;    #确保有成绩,0分也是有成绩,

9、查询「李」姓老师的数量

select count(tid) as '姓李的老师个数'
from teacher
where tname like '李%' ;

10、查询学过「张三」老师授课的同学的信息

思路：张三老师教的是哪门课程，有哪些学生选了张三老师教的那门课

select  student.*

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';

11、查询没有学全所有课程的同学的信息

#有一些同学一门课都没有选,所以成绩表SC里面没有这些学生的任何信息，
#现在题目的要求是没有学全，我的理解是SC表里的同学

select student.*
from sc join student on sc.sid=student.sid
group by sc.sid
having count(sc.cid)<(select count(course.cid)  from course);

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  distinct sid
from sc
where cid in (select cid from sc where sid=1) and sid!=1;

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

PS：使用了group_concat( )函数，将分组后指定字段的值连接起来，感觉使用时最好对连接的字段排序，以免出错

select student.*
from
(select sid, group_concat(cid order by cid) as  tt from sc  where sid=1  group by sid) a   #1号同学选修的课程
        #为了避免cid插入顺序引起的group_concat()结果有差异，所以对cid进行了排序
left join 

( select sid, group_concat(cid order by cid) as rr  from sc  where sid!=1  group by sid) b    on a.tt=b.rr
 #除1号同学外，其他同学选修的课程，将两张表进行连接，按照选修课程相同为条件进行连接

join student on b.sid=student.sid;

中间过程解析：

select sid, group_concat(cid order by cid) as  '1号同学选课情况' from sc  where sid=1  group by sid

运行结果：

select sid, group_concat(cid order by cid) as '除1号同学的选课情况'  from sc  where sid!=1  group by sid;

运行结果：

14、查询没学过"张三"老师讲授的任一门课程的学生姓名

#差集, SC表里的所有学生-选修了张三老师课程的学生,即为没有选修过张三老师任一课程的学生
select student.*
from
(select distinct sid from sc ) a  left join   #SC表全部的学号，去重

(select  sc.sid from ((sc left join course on sc.cid=course.cid) left join  teacher on course.tid=teacher.tid) where teacher.tname='张三')  b
 on a.sid=b.sid   

left join student on a.sid=student.sid

where b.sid is null;

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

PS：不明白它的这个平均成绩，假如1号同学选了4门课，其中有3门不及格，那么这个平均成绩是指这4门的，还是指这3门不及格的平均成绩

下面两种方法都可以选出不及格的同学，但是计算的平均成绩有所不同

一：不及格同学的所有选修课平均成绩

#这种方法计算的是每个同学所有选修课的平均成绩
select student.sid, student.sname, avg(sc.score) as '平均分'
from sc join student  on sc.sid=student.sid
group by sc.sid
having   sum(sc.score<60)>=2;  # 注意:使用的是sum()函数，不能使用count()函数

二：不及格同学的选修课中，不及格部分课程的平均成绩

#这种方法计算的是不及格同学中所选课程，不及格部分的平均成绩
select student.sid, student.sname, avg(sc.score) as '平均分'
from sc join student  on sc.sid=student.sid
where sc.score<60   # 先筛选出不及格的
group by sc.sid
having count(sc.cid)>=2;  

16、检索" 01 "课程分数小于 60，按分数降序排列的学生信息

select student.*
from sc join student on sc.sid=student.sid
where sc.cid=1 and sc.score<60
order by sc.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

A：当做总成绩处理，order by 排序时 平均成绩  这个字段绝对绝对不能加引号，加上引号之后就不能排序了

select sid, sum(score) as '总成绩', avg(score) as '平均成绩'
from sc
group by sid
order by  平均成绩  desc;

运行结果：真心搞不懂这个所有课程的成绩是指哪一种，这种按总成绩算好像没有什么意义，因为都知道平均成绩了

B：按每科成绩处理

select student.sid, sc.cid, sc.score, a.avgscore
from  student left  join  sc on student.sid=sc.sid
 left join
(select sid, avg(score) as  avgscore   from sc  group by sid)  a    on sc.sid=a.sid  #先计算每个同学的平均成绩
order by a.avgscore desc;

运行结果：图太大，分两张

18、查询各科成绩最高分、最低分和平均分：

select  cid, max(score) as '最高分',  min(score) as '最低分',  avg(score) as '平均分'
from sc
group by cid;

19、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

PS：round( )函数将小数保留2位，concat( )函数连接字符串

select  sc.cid, course.cname, max(sc.score) as '最高分',  min(sc.score) as '最低分', round(avg(sc.score),2) as '平均分',

concat(round(sum(case when sc.score>=60 and sc.score<70 then 1 else 0 end) /count(sc.sid)*100,2),'%')  as '及格率',
concat(round(sum(case when sc.score>=70 and sc.score<80 then 1 else 0 end)/count(sc.sid)*100,2),'%') as '中等率',
concat(round(sum(case when sc.score>=80 and sc.score<90 then 1 else 0 end)/count(sc.sid)*100,2),'%') as '优良率',
concat(round(sum(case when sc.score>=90 then 1 else 0 end )/count(sc.sid)*100,2),'%') as '优秀率'

from sc join course on sc.cid=course.cid
group by  sc.cid;

20、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

PS：order by子句中的 选修人数  这个字段一定不能加引号，否则order by子句排序的功能消失

select cid, count(sid) as '选修人数'
from sc
group by cid
order by 选修人数  desc,cid;

运行结果：

21、按各科成绩进行排序，并显示排名， Score 重复时保留名次空缺

PS：个人理解是，分课程，对学生进行排名，这道题目是名次不连续

select a.cid, a.sid, a.score , count(a.score<b.score)+1 as rank
from sc a left join sc b on a.cid=b.cid and a.score<b.score   #笛卡尔积连接, 然后筛选满足a.score<b.score的
group by a.cid, a.sid
order by a.cid, a.score desc;

运行结果：  两张图是一样的，右边的做了点标记

    

22、按各科成绩进行排序，并显示排名， Score 重复时合并名次

PS：个人理解，分课程，对学生排名，这道题目是名次要求连续

select a.cid, a.sid, a.score , count(distinct b.score)+1 as rank  #这里使用的是distinct b.score
from sc a left join sc b on a.cid=b.cid and a.score<b.score   #笛卡尔积连接, 然后筛选满足a.score<b.score的
group by a.cid, a.sid
order by a.cid, a.score desc;

运行结果：

          

23、查询各科成绩前三名的记录

PS：这道题目就是在21,22题的基础上添加一个having子句，筛选名次rank<=3，即前三名

A：名次不连续时的前三名

select a.cid, a.sid, a.score , count(a.score<b.score)+1 as rank
from sc a left join sc b on a.cid=b.cid and a.score<b.score   #笛卡尔积连接, 然后筛选满足a.score<b.score的
group by a.cid, a.sid
having rank<=3
order by a.cid, a.score desc;

运行结果：

B：名次连续时的前三名

select a.cid, a.sid, a.score , count(distinct b.score)+1 as rank  #这里使用的是distinct b.score
from sc a left join sc b on a.cid=b.cid and a.score<b.score   #笛卡尔积连接, 然后筛选满足a.score<b.score的
group by a.cid, a.sid
having rank<=3
order by a.cid, a.score desc;

运行结果：

24、查询学生的总成绩，并进行排名，总分重复时保留名次空缺

select a.sid, count(a.tt<b.rr)+1 as rank
from
(select sid, sum(score) as tt from sc group by sid ) a
left join
(select sid, sum(score) as rr from sc group by sid ) b   on  a.tt<b.rr
group by a.sid
order by rank;

运行结果：

25、 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 a.sid, count( distinct b.rr)+1 as rank
from
(select sid, sum(score) as tt from sc group by sid ) a
left join
(select sid, sum(score) as rr from sc group by sid ) b   on  a.tt<b.rr
group by a.sid
order by rank;

运行结果：

26、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, course.cname,count(sc.sid) as '人数',
concat(round(sum(case when score>=0 and score<60 then 1 else 0 end)/count(sc.sid)*100,2),'%') as '[0-60]',
concat(round(sum(case when score>=60 and score<70 then 1 else 0 end)/count(sc.sid)*100,2),'%') as '[60-70]',
concat(round((sum(case when score>=70 and score<85 then 1 else 0 end)/count(sc.sid))*100,2),'%') as '[70-85]',
concat(round((sum(case when score>=85 and score<100 then 1 else 0 end)/count(sc.sid))*100,2),'%') as '[85-100]'

from sc join course on sc.cid=course.cid
group by sc.cid;

运行结果：

select sc.cid, course.cname,count(sc.sid) as '该课程的总人数',
sum(case when score>=0 and score<60 then 1 else 0 end) as '[0-60]人数',
concat(round(sum(case when score>=0 and score<60 then 1 else 0 end)/count(sc.sid)*100,2),'%') as '[0-60]',
sum(case when score>=60 and score<70 then 1 else 0 end)  as '[60-70]人数',
concat(round(sum(case when score>=60 and score<70 then 1 else 0 end)/count(sc.sid)*100,2),'%') as '[60-70]',
sum(case when score>=70 and score<85 then 1 else 0 end)  as '[70-85]人数',
concat(round((sum(case when score>=70 and score<85 then 1 else 0 end)/count(sc.sid))*100,2),'%') as '[70-85]',
sum(case when score>=85 and score<100 then 1 else 0 end)  as '[85-100]人数',
concat(round((sum(case when score>=85 and score<100 then 1 else 0 end)/count(sc.sid))*100,2),'%') as '[85-100]'

from sc join course on sc.cid=course.cid
group by sc.cid;

运行结果：

27、查询每门课程被选修的学生数

select  cid, count(sid) as '学生数'
from sc
group by cid;

运行结果：

28、查询出只选修两门课程的学生学号和姓名

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

29、查询男生、女生人数

select ssex ,count(sid) as '人数'
from student
group by ssex;

30、查询名字中含有「风」字的学生信息

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

或使用正则表达式 REGEXP

select *
from student
where sname regexp  '风';

运行结果：

31、查询同名同性学生名单，并统计同名人数

#同性，不清楚是不是写错了

32、查询 1990 年出生的学生名单

select  sid, sname
from student
where  year(sborn)='';

33、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) as '平均成绩'
from sc
group  by cid
order by 平均成绩 desc, cid;

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

35、查询课程名称为「数学」，且分数低于 60 的学生姓名和分数

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

36、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 student.sid, sc.cid, sc.score
from student left join sc on student.sid=sc.sid;

37、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  student.sname, course.cname, sc.score
from sc join student on sc.sid=student.sid join course on sc.cid= course.cid
where sc.cid  in (select distinct sid from sc where score>70);

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 student.sid, student.sname
from sc join student on sc.sid=student.sid
where sc.cid=1 and sc.score>80;

39、求每门课程的学生人数

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

40、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

41、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

PS：这两道题，使用下面同样的代码都能运行出正确结果

select student.*, a.score
from sc a left join sc b on a.cid=b.cid and a.score<b.score   #对每门课程下的学生成绩排名

join course on a.cid=course.cid
join teacher on course.tid=teacher.tid and tname='张三'  #这一行和上一行，是为了筛选张三老师教的课
join student on a.sid=student.sid  # 获取学生信息

group by a.cid, a.sid
having count( a.score< b.score)=0   #筛选名次rank=1, count(a.score<b.score)+1=1，两端减去1，简化为count(a.score<b.score)=0

#上面这一行having 子句，这种格式也可以：having count( distinct b.score)=0 ,使用distinct
;

42、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

A：不同课程，成绩相同的学生信息

select sc.cid , sc.sid, sc.score
from sc join  (select cid, score from sc  group by cid, score having count(sid)>=2) a
        on sc.cid=a.cid and sc.score=a.score
order by cid ;

运行结果：

B：不同课程成绩相同的，学生信息

select    sc.sid, sc.cid , sc.score
from sc join  (select sid, score from sc  group by sid, score having count(cid)>=2) a
        on sc.sid=a.sid and sc.score=a.score
order by sid;

运行结果：

43、查询每门功成绩最好的前两名

A：保留名次空缺

select a.cid, a.sid,a.score , count(a.score<b.score)+1 as rank
from sc a left join sc b on a.cid=b.cid and a.score<b.score
group by a.cid,a.sid
having rank<=2
order by a.cid, a.score desc;

B：不保留名次空缺

select a.cid, a.sid,a.score , count(distinct b.score)+1 as rank
from sc a left join sc b on a.cid=b.cid and a.score<b.score
group by a.cid,a.sid
having rank<=2
order by a.cid, a.score desc;

44、统计每门课程的学生选修人数（超过 5 人的课程才统计）。

select  cid, count(sid) as '选修人数'
from sc
group by cid
having count(sid)>5;

45、检索至少选修两门课程的学生学号

select  sid
from sc
group by sid
having count(cid)>=2;

46、查询选修了全部课程的学生信息

select student.*
from sc join student on sc.sid=student.sid
group by  sc.sid
having count(cid)=(select count(cid) from  course);

47、查询各学生的年龄，只按年份来算

select  sid , sname, year(now())-year(sborn) as age
from student;

48、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 sid, sname ,sborn, timestampdiff(year,sborn,date_format(now(),'%Y-%m-%d')) as '题目要求计算的age',
year(now())-year(sborn) as '上一题的年龄'

from student;

运行结果：

49、查询本周过生日的学生

PS：

a.  在where子句中两端使用的函数要相同，因为week( )函数的周数是从0开始，weekofyear( )函数是从1开始

b. concat( )函数、concat_ws( )函数都可以，

c. 连接中间的年月日时，有两种形式：一种是使用函数date_format( )，另一种是整理出年、月、日，然后进行连接

d. 中间的连接符：-, /, #, * , $，这几个试了都可以

下面的where 子句除了最后一行，任选一行都能运行出正确的结果

select *
from student 

where weekofyear(concat(year(now()),'-',date_format(sborn,'%m-%d')))=weekofyear(now());  #成功

where weekofyear(concat(year(now()),'/',month(sborn),'/',day(sborn)))=weekofyear(now());

where week(concat(year(now()),'-',month(sborn),'-',day(sborn)))=week(now());

where week(concat_ws('',year(now()),month(sborn),day(sborn)))=week(now());

where week(concat_ws('-',year(now()),date_format(sborn,'%m-%d')))=week(now());

#where  weekofyear(year(now())&"-"&month(sborn)&"-"&day(sborn))=weekofyear(now()); #这种形式的连接报错

50、查询下周过生日的学生

select *
from student
where week(concat_ws('-',year(now()),date_format(sborn,'%m-%d')))=week(now())+1;

51、查询本月过生日的学生

select  sid , sname, sborn
from student
where month(sborn)=month(now());

52、查询下月过生日的学生

select  sid , sname, sborn
from student
where month(sborn)=month(now())+1;

做这些题目时，感觉自己语文水平有待提升，题目都读不懂了