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MySQL数据库----多表查询
阅读量:6977 次
发布时间:2019-06-27

本文共 8605 字,大约阅读时间需要 28 分钟。

一、介绍

首先先准备表

员工表和部门表

#建表create table department(id int,name varchar(20) );create table employee1(id int primary key auto_increment,name varchar(20),sex enum('male','female') not null default 'male',age int,dep_id int);#插入数据insert into department values(200,'技术'),(201,'人力资源'),(202,'销售'),(203,'运营');insert into employee1(name,sex,age,dep_id) values('egon','male',18,200),('alex','female',48,201),('wupeiqi','male',38,201),('yuanhao','female',28,202),('liwenzhou','male',18,200),('jingliyang','female',18,204);

查看表:

二、多表连接查询

1.交叉连接:不适用任何匹配条件。生成笛卡尔积、

select * from employee1 ,department;

2.内连接:找两张表共有的部分,相当于利用条件从笛卡尔积结果中筛选出了正确的结果。(只连接匹配的行)

#找两张表共有的部分,相当于利用条件从笛卡尔积结果中筛选出了正确的结果#department没有204这个部门,因而employee表中关于204这条员工信息没有匹配出来select * from employee1,department where employee1.dep_id=department.id;#上面用where表示的可以用下面的内连接表示,建议使用下面的那种方法select * from employee1 inner join department on employee1.dep_id=department.id; #也可以这样表示哈
select employee1.id,employee1.name,employee1.age,employee1.sex,department.name from employee1,department where employee1.dep_id=department.id;

 

注意:内连接的join可以忽略不写,但是还是加上看起来清楚点

3.左连接:优先显示左表全部记录。

#左链接:在按照on的条件取到两张表共同部分的基础上,保留左表的记录select * from employee1 left join department on department.id=employee1.dep_id; select * from department left join  employee1 on department.id=employee1.dep_id;

 

4.右链接:优先显示右表全部记录。

#右链接:在按照on的条件取到两张表共同部分的基础上,保留右表的记录select * from employee1 right join department on department.id=employee1.dep_id;
select * from department right join employee1 on department.id=employee1.dep_id;

5.全外连接:显示左右两个表的全部记录。

注意:mysql不支持全外连接 full join强调:mysql可以使用union间接实现全外连接
select * from employee1 left join department on department.id=employee1.dep_idunionselect * from employee1 right join department on department.id=employee1.dep_id; 

三、符合条件连接查询

示例1:以内连接的方式查询employee和department表,并且employee表中的age字段值必须大于25, 即找出公司所有部门中年龄大于25岁的员工
select * from employee1 inner join department on employee1.dep_id=department.id and age>25;

 

示例2:以内连接的方式查询employee和department表,并且以age字段的升序方式显示
select * from employee1 inner join department on employee1.dep_id=department.id = and age>25 and age>25 order by age asc;

四、子查询

#1:子查询是将一个查询语句嵌套在另一个查询语句中。#2:内层查询语句的查询结果,可以为外层查询语句提供查询条件。#3:子查询中可以包含:IN、NOT IN、ANY、ALL、EXISTS 和 NOT EXISTS等关键字#4:还可以包含比较运算符:= 、 !=、> 、

小练习

#查询平均年龄在25岁以上的部门名select name from department where id in (select dep_id from employee group by dep_id having avg(age) > 25);#查看技术部员工姓名select name from employee where dep_id = (select id from department where name='技术');#查看小于2人的部门名select name from department where id in (select dep_id from employee1 group by dep_id having count(id) < 2)unionselect name from department where id not in (select distinct dep_id from employee1);#提取空部门                              #有人的部门select * from department where id not in (select distinct dep_id from employee1);或者:select name from department where id in(select dep_id from employee1 group by dep_id having count(id) < 2unionselect id from department where id not in (select distinct dep_id from employee1););

五、综合练习

一、SELECT语句关键字的定义顺序

SELECT DISTINCT 
FROM
JOIN
ON
WHERE
GROUP BY
HAVING
ORDER BY
LIMIT
SELECT语句关键字的定义顺序

二 、SELECT语句关键字的执行顺序

(7)     SELECT (8)     DISTINCT 
(1) FROM
(3)
JOIN
(2) ON
(4) WHERE
(5) GROUP BY
(6) HAVING
(9) ORDER BY
(10) LIMIT

init.sql文件内容

 

 从init.sql文件中导入数据

#准备表、记录mysql> create database db1;mysql> use db1;mysql> source /root/init.sql
1、查询所有的课程的名称以及对应的任课老师姓名2、查询学生表中男女生各有多少人3、查询物理成绩等于100的学生的姓名4、查询平均成绩大于八十分的同学的姓名和平均成绩5、查询所有学生的学号,姓名,选课数,总成绩6、 查询姓李老师的个数7、 查询没有报李平老师课的学生姓名8、 查询物理课程比生物课程高的学生的学号9、 查询没有同时选修物理课程和体育课程的学生姓名10、查询挂科超过两门(包括两门)的学生姓名和班级、查询选修了所有课程的学生姓名12、查询李平老师教的课程的所有成绩记录 13、查询全部学生都选修了的课程号和课程名14、查询每门课程被选修的次数15、查询之选修了一门课程的学生姓名和学号16、查询所有学生考出的成绩并按从高到低排序(成绩去重)17、查询平均成绩大于85的学生姓名和平均成绩18、查询生物成绩不及格的学生姓名和对应生物分数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20、查询每门课程成绩最好的前两名学生姓名21、查询不同课程但成绩相同的学号,课程号,成绩22、查询没学过“叶平”老师课程的学生姓名以及选修的课程名称;23、查询所有选修了学号为1的同学选修过的一门或者多门课程的同学学号和姓名;24、任课最多的老师中学生单科成绩最高的学生姓名
题目

答案待续。。。。

1、查询所有的课程的名称以及对应的任课老师姓名select cname 课程名称,tname 老师姓名 from course  left join teacher on course.teacher_id = teacher.tid;2、查询学生表中男女生各有多少人select gender ,count(gender) from student group by gender;3、查询物理成绩等于100的学生的姓名select sname from student inner join score on student.sid = score.student_idjoin course on score.course_id = course.cidwhere cname = '物理' and 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(1):select sname 姓名,avg(num) 平均成绩 from student inner join score on student.sid = score.student_id group by student_id having avg(num)>80;(2):select sname,avg(num) from student inner join score on student.sid = score.student_id group by student_id having avg(num)>805、查询所有学生的学号,姓名,选课数,总成绩select student.sid 学号,sname 姓名,count(course_id) 选课数,sum(num) 总成绩 from student left join score on student.sid = score.student_idgroup by student.sid ;6、 查询姓李老师的个数select count(tid) from teacher where tname like '李%';7、 查询没有报李平老师课的学生姓名select sname 姓名 from student where sid not in (select student_id from score where course_id in(select cid from course where teacher_id =(select tid from teacher where tname = '李平老师')));8、 查询物理课程比生物课程高的学生的学号select t1.student_id  学号 from (select student_id ,num from score inner join course on score.course_id=course.cid  where cname='物理' )as t1inner join (select student_id , num from score inner join course on score.course_id=course.cid  where cname = '生物') as t2on t1.student_id = t2.student_idwhere 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):select sname from student inner join score on student.sid = score.student_idjoin course on course.cid=score.course_id and cname in ('物理','体育') group by student_id having count(course_id)!=2;(2):select sname from student join score  on student.sid = score.student_idjoin course on course.cid=score.course_id where cname ='物理' or cname= '体育'group by student_id having count(course_id) !=2;10、查询挂科超过两门(包括两门)的学生姓名和班级select sname 姓名,caption 班级 from student inner join score on student.sid = score.course_idjoin class on class.cid = score.course_idwhere num<60 group by student_id having count(course_id)>=2;11、查询选修了所有课程的学生姓名(1)select sname 姓名 ,所有的课程数 from student inner join (select student_id,count(course_id) 所有的课程数 from score group by student_id having count(course_id) = (select count(cid) from course)) as t1on t1.student_id = student.sid;(2)select sname,count(course_id) from student inner join score on student.sid = score.student_idgroup by student_id having count(course_id)=(select count(cid) from course);12、查询李平老师教的课程的所有成绩记录(1):select num from score inner join course on course.cid=score.course_idjoin teacher on teacher.tid=course.teacher_idwhere tname = '李平老师';(2):select num from score where course_id in(select cid from course where teacher_id=(select tid from teacher where tname='李平老师'));13、查询全部学生都选修了的课程号和课程名select cid 课程号,cname 课程名 from courseselect ;14、查询每门课程被选修的次数 (1)select course.cname,count(student_id) 选课人数 from score inner join course on score.course_id=course.cidgroup by course_id;(2):也可以按照name分组select course.cname,count(student_id) 选课人数 from score inner join course on score.course_id=course.cidgroup by cname;15、查询之选修了一门课程的学生姓名和学号select sname 姓名,student_id 学号 from student inner join score on student.sid = score.student_idgroup by student_id having count(course_id)=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)select distinct num from score order by num desc;17、查询平均成绩大于85的学生姓名和平均成绩(1):select sname 姓名,avg(num) 平均成绩 from student inner join score on student.sid = score.student_idgroup by student_id having avg(num)>85;(2):select student.sname,avg_num from student inner join(select student_id,avg(num) as avg_num from score group by student_id having avg(num) > 85) t1on student.sid=t1.student_id;18、查询生物成绩不及格的学生姓名和对应生物分数(1):select student.sname ,num 生物成绩 from student inner join score on student.sid = score.student_idjoin course on course.cid=score.course_idwhere cname='生物' and num<60;(2):select student.sname,t1.num from student inner join(select student_id,num from scorewhere course_id=(select cid from course where cname='生物') and num < 60) t1on t1.student_id=student.sid;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select sname from student where sid=(select student_id from score where course_id in (select cid from course where teacher_id=(select tid from teacher where tname='李平老师')) group by student_id order by avg(num) desc limit 1)
答案

转载于:https://www.cnblogs.com/TheLand/p/8481777.html

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