day41:MYSQL:select查询练习题

目录

1.表结构

2.创建表和插入数据

3.习题

1.表结构

 2.建表和插入数据

# 创建班级表
create table class(
cid int primary key auto_increment,
caption varchar(32) not null
);

# 创建学生表
create table student(
sid int primary key auto_increment,
gender char(1) not null,
class_id int not null,
sname varchar(32) not null,
foreign key(class_id) references class(cid) on delete cascade on update cascade
);

# 创建老师表
create table teacher(
tid int primary key auto_increment,
tname varchar(32) not null
);

# 创建课程表
create table course(
cid int primary key auto_increment,
cname varchar(32) not null,
teacher_id int not null,
foreign key(teacher_id) references teacher(tid) on delete cascade on update cascade
);

# 创建成绩表
create table score(
sid int primary key auto_increment,
student_id int not null,
course_id int not null,
num int not null,
foreign key(student_id) references student(sid) on delete cascade on update cascade,
foreign key(course_id) references course(cid) on delete cascade on update cascade
);


# 班级表插入记录
insert into class values
('1', '三年二班'), 
('2', '三年三班'), 
('3', '一年二班'), 
('4', '二年一班');

# 学生表插入记录
insert into student values
('1', '男', '1', '理解'), 
('2', '女', '1', '钢蛋'), 
('3', '男', '1', '张三'), 
('4', '男', '1', '张一'), 
('5', '女', '1', '张二'), 
('6', '男', '1', '张四'), 
('7', '女', '2', '铁锤'),
('8', '男', '2', '李三'), 
('9', '男', '2', '李一'), 
('10', '女', '2', '李二'), 
('11', '男', '2', '李四'), 
('12', '女', '3', '如花'), 
('13', '男', '3', '刘三'), 
('14', '男', '3', '刘一'), 
('15', '女', '3', '刘二'), 
('16', '男', '3', '刘四');

# 老师表插入记录
insert into teacher values
('1', '张磊'), 
('2', '李平'), 
('3', '刘海燕'), 
('4', '朱云海'), 
('5', '李春秋');

# 课程表插入记录
insert into course values
('1', '生物', '1'), 
('2', '物理', '2'), 
('3', '体育', '3'), 
('4', '美术', '2');

# 成绩表插入记录
insert into score values
('1', '1', '1', '10'), 
('2', '1', '2', '9'), 
('3', '1', '3', '76'),
('5', '1', '4', '66'), 
('6', '2', '1', '8'), 
('8', '2', '3', '68'), 
('9', '2', '4', '99'), 
('10', '3', '1', '77'), 
('11', '3', '2', '66'), 
('12', '3', '3', '87'), 
('13', '3', '4', '99'), 
('14', '4', '1', '79'), 
('15', '4', '2', '11'), 
('16', '4', '3', '67'), 
('17', '4', '4', '100'), 
('18', '5', '1', '79'), 
('19', '5', '2', '11'), 
('20', '5', '3', '67'), 
('21', '5', '4', '100'), 
('22', '6', '1', '9'), 
('23', '6', '2', '100'), 
('24', '6', '3', '67'), 
('25', '6', '4', '100'), 
('26', '7', '1', '9'), 
('27', '7', '2', '100'), 
('28', '7', '3', '67'), 
('29', '7', '4', '88'), 
('30', '8', '1', '9'), 
('31', '8', '2', '100'), 
('32', '8', '3', '67'),
('33', '8', '4', '88'), 
('34', '9', '1', '91'), 
('35', '9', '2', '88'), 
('36', '9', '3', '67'), 
('37', '9', '4', '22'), 
('38', '10', '1', '90'), 
('39', '10', '2', '77'), 
('40', '10', '3', '43'), 
('41', '10', '4', '87'), 
('42', '11', '1', '90'), 
('43', '11', '2', '77'), 
('44', '11', '3', '43'), 
('45', '11', '4', '87'), 
('46', '12', '1', '90'), 
('47', '12', '2', '77'), 
('48', '12', '3', '43'), 
('49', '12', '4', '87'), 
('52', '13', '3', '87');

3.习题

1、查询所有的课程的名称以及对应的任课老师姓名

# 1、查询所有的课程的名称以及对应的任课老师姓名
# where
select 
    course.cname,teacher.tname
from 
    teacher,course
where 
    teacher.tid = course.teacher_id
    
# inner join
select 
    course.cname,teacher.tname
from 
    teacher inner join course on teacher.tid = course.teacher_id

2、查询学生表中男女生各有多少人

# 2、查询学生表中男女生各有多少人
select 
    gender,count(*)
from 
    student
group by
    gender

3、查询物理成绩等于100的学生的姓名

# 3、查询物理成绩等于100的学生的姓名
# where
select 
    student.sname , score.num
from 
    score,student,course
where
    score.student_id = student.sid 
    and
    score.course_id = course.cid
    and
    course.cname = "物理"
    and
    score.num = 100
    
# inner join 
select 
    student.sname , score.num
from 
    score inner join student on  score.student_id = student.sid 
    inner join course on score.course_id = course.cid
where
    course.cname = "物理"
    and
    score.num = 100

4、查询平均成绩大于八十分的同学的姓名和平均成绩

# where 写法
select
    student_id,student.sname,avg(num)
from
    score,student
where 
    score.student_id = student.sid
group by
    student_id
having
    avg(num) > 80

# inner join
select
    student_id,student.sname,avg(num)
from
    score inner join student on score.student_id = student.sid
group by
    student_id
having
    avg(num) > 80

5、查询所有学生的学号，姓名，选课数，总成绩3

# 选课数
select
    student_id,count(*)
from 
    score
group by 
    student_id
    
# 总成绩
select 
    student_id,sum(num)
from 
    score
group by
    student_id

# where
select 
    student_id,sname,count(*),sum(num)
from 
    score,student
where
    score.student_id = student.sid
group by
    student_id

# inner join 
select 
    student_id,sname,count(*),sum(num)
from 
    score inner join student on score.student_id = student.sid
group by
    student_id

# ( 附加所有学生 )
# right join 
select 
    student.sid,sname,count(score.course_id),sum(num)
from 
    score right join student on score.student_id = student.sid
group by
    student.sid

# left join 
select 
    student.sid,sname,count(score.course_id),sum(num)
from 
    student left join score  on score.student_id = student.sid
group by
    student.sid

6、 查询姓李老师的个数

# 6、 查询姓李老师的个数
select
    count(*)
from 
    teacher
where 
    tname like '李%'

7、 查询没有报李平老师课的学生姓名

# 7、 查询没有报李平老师课的学生姓名
# 1.报了李平老师课程的学生id是?
"""distinct 去重  
distinct student_id    ok
distinct(student_id)   ok
"""
select 
    distinct( student_id )
from 
    teacher,course,score
where
    teacher.tid = course.teacher_id
    and
    course.cid = score.course_id
    and 
    teacher.tname = "李平"

# 2.查询学生表,除了这个id的剩下的就是没有报李平老师课程的
select 
    student.sname
from 
    student
where 
    sid not in (1号数据)

#3.综合拼接
select
    student.sname
from 
    student
where 
    sid not in (select 
    distinct( student_id )
from 
    teacher,course,score
where
    teacher.tid = course.teacher_id
    and
    course.cid = score.course_id
    and 
    teacher.tname = "李平")

8、 查询物理课程的分数比生物课程的分数高的学生的学号

# 8、 查询物理课程的分数比生物课程的分数高的学生的学号

# 1.物理课程学生分数
select 
    score.student_id as t1_id , score.num as num , course.cid ,course.cname
from 
    course inner join score on course.cid = score.course_id
where
    course.cname = "物理"
    
# 2.生物课程学生分数
select 
    score.student_id as t2_id , score.num as num , course.cid ,course.cname
from 
    course inner join score on course.cid = score.course_id
where
    course.cname = "生物"
    
    
# 综合拼接
"""
# 格式
select 
    t1.t1_id
from 
    (1) inner join (2) on 1.student_id = 2.student_id
where
    1.num > 2.num
"""
select 
    t1.t1_id
from 
    (select 
    score.student_id as t1_id , score.num as num , course.cid ,course.cname
from 
    course inner join score on course.cid = score.course_id
where
    course.cname = "物理") as t1 inner join (select 
    score.student_id as t2_id , score.num as num , course.cid ,course.cname
from 
    course inner join score on course.cid = score.course_id
where
    course.cname = "生物") as t2 on t1.t1_id = t2.t2_id
where
    t1.num > t2.num

9、 查询没有同时选修物理课程和体育课程的学生姓名

# 1.找物理和体育的课程id
select 
    cid
from 
    course
where
    cname = "物理" or cname = "体育"
    
# 2.找学习体育物理课程的学生id 
select 
    student_id
from 
    score
where 
    course_id in (2,3)

# 拼装数据
select 
    student_id
from 
    score
where 
    course_id in (select 
    cid
from 
    course
where
    cname = "物理" or cname = "体育")

# 3.(同时)学习体育物理课程的学生id
select 
    student_id
from 
    score
where 
    course_id in (select 
    cid
from 
    course
where
    cname = "物理" or cname = "体育")

group by
    score.student_id
having 
    count(*) = 2

# 4.除了通过学习物理和体育的学生之外,剩下的都是没有同时学习的学生id
select     
    sid,sname
from 
    student
where 
    sid not in (3号)
    
# 综合拼装:
select     
    sid,sname
from 
    student
where 
    sid not in (select 
    student_id
from 
    score
where 
    course_id in (select 
    cid
from 
    course
where
    cname = "物理" or cname = "体育")

group by
    score.student_id
having 
    count(*) = 2)

10、查询挂科超过两门(包括两门)的学生姓名和班级

# 10、查询挂科超过两门(包括两门)的学生姓名和班级
"""挂科<60"""
select
    student_id,student.sname,class.caption
from 
    score inner join student on score.student_id = student.sid
    inner join class on class.cid = student.class_id
where 
    num < 60
group by
    student_id
having 
    count(*) >= 2

11、查询选修了所有课程的学生姓名

# 11、查询选修了所有课程的学生姓名
# 1.统计所有课程总数
select count(*) from course

# 2.按照学生分类,总数量是1号查询出来的数据,等价于学了所有课程
select 
    student.sid, student.sname
from
    score inner join student on  score.student_id = student.sid
group by 
    score.student_id
having 
    count(*) = (1号)
    
# 综合拼接
select 
    student.sid, student.sname
from
    score inner join student on  score.student_id = student.sid
group by 
    score.student_id
having 
    count(*) = (select count(*) from course)

12、查询李平老师教的课程的所有成绩记录

# 12、查询李平老师教的课程的所有成绩记录
# 内联
select
    score.student_id , course.cname , score.num
from 
    teacher , course , score
where
    teacher.tid = course.teacher_id
    and
    score.course_id = course.cid
    and 
    teacher.tname = "李平"

# 子查询
# 1.找李平老师的课程id
select 
    course.cid
from 
    teacher,course
where
    teacher.tid = course.teacher_id
    and
    teacher.tname = "李平"
# 2.通过课程id号 找score里面的数据
select *
from 
    score
where 
    course_id in (1号)

# 综合拼接
select 
    score.student_id ,score.num
from 
    score
where 
    course_id in (select 
    course.cid
from 
    teacher,course
where
    teacher.tid = course.teacher_id
    and
    teacher.tname = "李平")

13、查询全部学生都选修了的课程号和课程名

# 13、查询全部学生都选修了的课程号和课程名
# 1.通过score表,找有成绩的学生个数
select
    count(distinct student_id)
from
    score

# 2.按照课程分类,筛选学生个数等于13的课程id
select
    course_id
from 
    score
group by
    course_id
having 
    count(*) = 13
    
# 综合拼接
select
    course_id,course.cname
from 
    score,course
where
    score.course_id = course.cid
group by
    course_id
having 
    count(*) = (select
    count(distinct student_id)
from
    score)

14、查询每门课程被选修的次数

# 14、查询每门课程被选修的次数;
select 
    course_id,count(*)
from 
    score
group by 
    course_id

15、查询只选修了一门课程的学生学号和姓名

# 15、查询只选修了一门课程的学生学号和姓名
# 1.按照学生分类,统计课程个数为1
select 
    student_id
from 
    score
group by 
    student_id
having
    count(*) = 1

# 2.顺手连带一个学生表student , 通过id拿姓名

select 
    student_id,student.sname
from 
    score inner join student on score.student_id = student.sid
group by 
    student_id
having
    count(*) = 1

16、查询所有学生考出的成绩并按从高到低排序（成绩去重）

# 16、查询所有学生考出的成绩并按从高到低排序（成绩去重）
select 
    distinct num,group_concat(student_id)
from
    score
group by
    num
order by
    num desc

#  其他同学想法
select 
    avg(num),sum(num)
from 
    score
group by
    student_id
order by 
    avg(num) desc 

17、查询平均成绩大于85的学生姓名和平均成绩

# 17、查询平均成绩大于85的学生姓名和平均成绩
# 子查询

# 1.找学生id
select 
    student_id
from 
    score
group by
    student_id
having 
    avg(num) > 85
    
# 2.找学生表对应数据
select sname 
from 
    student
where     
    id = (1号)

# 综合拼接
select 
    sid,sname 
from 
    student
where     
    sid = (select 
    student_id
from 
    score
group by
    student_id
having 
    avg(num) > 85)

18、查询生物成绩不及格的学生姓名和对应生物分数

# 18、查询生物成绩不及格的学生姓名和对应生物分数
select 
    student.sname,score.num,course.cname
from 
    course inner join score on course.cid = score.course_id
    inner join student on student.sid = score.student_id
where
    score.num < 60
    and
    course.cname = "生物"

19、查询在所有选修了李平老师课程的学生中，这些课程(李平老师的课程，不是所有课程)平均成绩最高的学生姓名

# 19、查询在所有选修了李平老师课程的学生中，这些课程(李平老师的课程，不是所有课程)平均成绩最高的学生姓名
# 1.找李平老师教的课程id
select 
    course.cid
from 
    teacher , course
where 
    teacher.tid = course.teacher_id
    and
    teacher.tname = "李平" 
# 2 4

# 2.学习李平老师课程的学生中,按照学生分类,找平均分最高的id
select
    score.student_id , avg(num)
from 
    score
where
    score.course_id in (2,4)
group by
    score.student_id
order by 
    avg(num) desc limit 1

# 3.通过学生id 顺带着连一张student学生表,找出姓名
select
    score.student_id , student.sname  , avg(num) 
from 
    score,student
where
    score.student_id = student.sid
    and
    score.course_id in (1号)
group by
    score.student_id
order by 
    avg(num) desc limit 1
    
    
# 4.综合拼接
select
    score.student_id , student.sname  , avg(num) 
from 
    score,student
where
    score.student_id = student.sid
    and
    score.course_id in (select 
    course.cid
from 
    teacher , course
where 
    teacher.tid = course.teacher_id
    and
    teacher.tname = "李平" )
group by
    score.student_id
order by 
    avg(num) desc limit 1

20、查询每门课程成绩最好的课程id、学生姓名和分数

# 20、查询每门课程成绩最好的课程id、学生姓名和分数
# 1.找分数最大值.
select 
    course_id,max(num) as max_num
from 
    score
group by
    score.course_id

# 2.找出该分数对应的那批学生

select 
    *
from 
    score as t1 inner join student as t2 on t1.student_id  = t2.sid
    inner join (1号) as t3 on  t1.course_id = t3.course_id
    

# 综合拼接
select 
    t1.course_id , t2.sname , t3.max_num
from 
    score as t1 inner join student as t2 on t1.student_id  = t2.sid
    inner join (select 
    course_id,max(num) as max_num
from 
    score
group by
    score.course_id) as t3 on  t1.course_id = t3.course_id

where
    t1.num = t3.max_num

21、查询不同课程但成绩相同的课程号、学生号、成绩

# 21、查询不同课程但成绩相同的课程号、学生号、成绩 
"""不同的课程 如果使用!= 相同的数据返回来又查询了一遍,翻倍,为了防止翻倍重复查询使用>或者< """
select 
    s1.student_id as s1_sid,
    s2.student_id as s2_sid,
    s1.course_id as s1_cid,
    s2.course_id as s2_cid,
    s1.num as s1_num,
    s2.num as s2_num
from 
    score as s1,
    score as s2
where     
    s1.course_id > s2.course_id
    and
    s1.num = s2.num

22、查询没学过“李平”老师课程的学生姓名以及选修的课程名称

和第7题重复了!!

23、查询所有选修了学号为2的同学选修过的一门或者多门课程的同学学号和姓名

# 23、查询所有选修了学号为2的同学选修过的一门或者多门课程的同学学号和姓名 
# 1.学号为2的学生,选了什么学科
select 
    course_id
from 
    score
where
    student_id = 2  # 1 3 4
    
# 2.学过1 3 4 学科的学生都有谁
select 
    distinct student_id,student.sname
from
    score inner join student  on score.student_id= student.sid
where 
    course_id in (1,3,4)
    
# 综合拼接
select 
    distinct student_id,student.sname
from
    score inner join student  on score.student_id= student.sid
where 
    course_id in (select 
    course_id
from 
    score
where
    student_id = 2)

24、任课最多的老师中学生单科成绩最高的课程id、学生姓名和分数

# 24、任课最多的老师中学生单科成绩最高的课程id、学生姓名和分数

# 1.老师任何的最大数量是多少
"""任课数量为2的老师可能不止一个"""
select 
    count(*)
from    
    course
group by
    teacher_id
order by 
    count(*) desc limit 1
    
# 2.找最大任课数量为2的老师id
select 
    teacher_id
from 
    course
group by
    teacher_id
having 
    count(*) = (1号)
    
# 综合拼接
select 
    teacher_id
from 
    course
group by
    teacher_id
having 
    count(*) = (select 
    count(*)
from    
    course
group by
    teacher_id
order by 
    count(*) desc limit 1)


# 3.通过老师id 找课程
select 
    cid
from 
    course
where
    teacher_id in (2号)


# 综合拼接
# 2 4 
select 
    cid
from 
    course
where
    teacher_id in (select 
    teacher_id
from 
    course
group by
    teacher_id
having 
    count(*) = (select 
    count(*)
from    
    course
group by
    teacher_id
order by 
    count(*) desc limit 1))

# 4.通过该课程号,找其中的最大值(最大分数)
select 
    course_id,
    max(num) as max_num
from 
    score
where
    course_id in (3号)
group by
    course_id
    
# 综合拼接
select
    course_id,
    max(num) as max_num
from 
    score
where
    course_id in (select 
    cid
from 
    course
where
    teacher_id in (select 
    teacher_id
from 
    course
group by
    teacher_id
having 
    count(*) = (select 
    count(*)
from    
    course
group by
    teacher_id
order by 
    count(*) desc limit 1)))
group by
    course_id


# 5.把对应的学生姓名,最大分数拼在一起,做一次单表查询

select 
    *
from 
    score as t1 inner join student as t2 on  t1.student_id = t2.sid
    inner join (4号) as t3 on t3.course_id = t1.course_id

# 综合拼接
select 
    *
from 
    score as t1 inner join student as t2 on  t1.student_id = t2.sid
    inner join (select
    course_id,
    max(num) as max_num
from 
    score
where
    course_id in (select 
    cid
from 
    course
where
    teacher_id in (select 
    teacher_id
from 
    course
group by
    teacher_id
having 
    count(*) = (select 
    count(*)
from    
    course
group by
    teacher_id
order by 
    count(*) desc limit 1)))
group by
    course_id) as t3 on t3.course_id = t1.course_id


# 6.把分数是100分的最大值的学员查出来
select 
    *
from 
    score as t1 inner join student as t2 on  t1.student_id = t2.sid
    inner join (4号) as t3 on t3.course_id = t1.course_id
where
    t1.num = t3.max_num
    

# 综合拼接
select 
    t1.course_id,t2.sname,t3.max_num
from 
    score as t1 inner join student as t2 on  t1.student_id = t2.sid
    inner join (select
    course_id,
    max(num) as max_num
from 
    score
where
    course_id in (select 
    cid
from 
    course
where
    teacher_id in (select 
    teacher_id
from 
    course
group by
    teacher_id
having 
    count(*) = (select 
    count(*)
from    
    course
group by
    teacher_id
order by 
    count(*) desc limit 1)))
group by
    course_id) as t3 on t3.course_id = t1.course_id
where
    t1.num = t3.max_num