day41:MYSQL:select查询练习题
- 2020 年 8 月 31 日
- 筆記
- PythonS31-笔记, Python全栈31期-笔记
目录
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