这篇文章主要介绍了MySQL多表查询,结合实例形式详细分析了mysql多表查询中的链接查询、子查询等相关操作技巧与使用注意事项,需要的朋友可以参考下
本文实例讲述了MySQL多表查询。分享给大家供大家参考，具体如下：
准备工作：准备两张表，部门表（department）、员工表（employee）
create table department( id int, name varchar(20) ); create table employee( 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 employee(name,sex,age,dep_id) values ('egon','male',18,200), ('alex','female',48,201), ('wupeiqi','male',38,201), ('yuanhao','female',28,202), ('nvshen','male',18,200), ('xiaomage','female',18,204) ; # 查看表结构和数据 mysql> desc department; +-------+-------------+------+-----+---------+-------+ | Field | Type | Null | Key | Default | Extra | +-------+-------------+------+-----+---------+-------+ | id | int(11) | YES | | NULL | | | name | varchar(20) | YES | | NULL | | +-------+-------------+------+-----+---------+-------+ 2 rows in set (0.19 sec)
mysql> desc employee; +--------+-----------------------+------+-----+---------+----------------+ | Field | Type | Null | Key | Default | Extra | +--------+-----------------------+------+-----+---------+----------------+ | id | int(11) | NO | PRI | NULL | auto_increment | | name | varchar(20) | YES | | NULL | | | sex | enum('male','female') | NO | | male | | | age | int(11) | YES | | NULL | | | dep_id | int(11) | YES | | NULL | | +--------+-----------------------+------+-----+---------+----------------+ 5 rows in set (0.01 sec) mysql> select * from department; +------+--------------+ | id | name | +------+--------------+ | 200 | 技术 | | 201 | 人力资源 | | 202 | 销售 | | 203 | 运营 | +------+--------------+ 4 rows in set (0.02 sec)
mysql> select * from employee; +----+----------+--------+------+--------+ | id | name | sex | age | dep_id | +----+----------+--------+------+--------+ | 1 | egon | male | 18 | 200 | | 2 | alex | female | 48 | 201 | | 3 | wupeiqi | male | 38 | 201 | | 4 | yuanhao | female | 28 | 202 | | 5 | nvshen | male | 18 | 200 | | 6 | xiaomage | female | 18 | 204 | +----+----------+--------+------+--------+ 6 rows in set (0.00 sec)
ps:观察两张表，发现department表中id=203部门在employee中没有对应的员工，发现employee中id=6的员工在department表中没有对应关系。
一多表链接查询
SELECT 字段列表
FROM 表1 INNER|LEFT|RIGHT JOIN 表2
ON 表1.字段 = 表2.字段;
（1）先看第一种情况交叉连接：不适用任何匹配条件。生成笛卡尔积.--->重复最多
mysql> select * from employee,department; +----+----------+--------+------+--------+------+--------------+ | id | name | sex | age | dep_id | id | name | +----+----------+--------+------+--------+------+--------------+ | 1 | egon | male | 18 | 200 | 200 | 技术 | | 1 | egon | male | 18 | 200 | 201 | 人力资源 | | 1 | egon | male | 18 | 200 | 202 | 销售 | | 1 | egon | male | 18 | 200 | 203 | 运营 | | 2 | alex | female | 48 | 201 | 200 | 技术 | | 2 | alex | female | 48 | 201 | 201 | 人力资源 | | 2 | alex | female | 48 | 201 | 202 | 销售 | | 2 | alex | female | 48 | 201 | 203 | 运营 | | 3 | wupeiqi | male | 38 | 201 | 200 | 技术 | | 3 | wupeiqi | male | 38 | 201 | 201 | 人力资源 | | 3 | wupeiqi | male | 38 | 201 | 202 | 销售 | | 3 | wupeiqi | male | 38 | 201 | 203 | 运营 | | 4 | yuanhao | female | 28 | 202 | 200 | 技术 | | 4 | yuanhao | female | 28 | 202 | 201 | 人力资源 | | 4 | yuanhao | female | 28 | 202 | 202 | 销售 | | 4 | yuanhao | female | 28 | 202 | 203 | 运营 | | 5 | nvshen | male | 18 | 200 | 200 | 技术 | | 5 | nvshen | male | 18 | 200 | 201 | 人力资源 | | 5 | nvshen | male | 18 | 200 | 202 | 销售 | | 5 | nvshen | male | 18 | 200 | 203 | 运营 | | 6 | xiaomage | female | 18 | 204 | 200 | 技术 | | 6 | xiaomage | female | 18 | 204 | 201 | 人力资源 | | 6 | xiaomage | female | 18 | 204 | 202 | 销售 | | 6 | xiaomage | female | 18 | 204 | 203 | 运营 |
（2）内连接：只连接匹配的行，以双方为基准
#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匹配的结果 #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 mysql> select employee.id,employee.name,employee.age,employee.sex,department.name from employee inner join department on employee.dep_id=department.id; +----+---------+------+--------+--------------+ | id | name | age | sex | name | +----+---------+------+--------+--------------+ | 1 | egon | 18 | male | 技术 | | 2 | alex | 48 | female | 人力资源 | | 3 | wupeiqi | 38 | male | 人力资源 | | 4 | yuanhao | 28 | female | 销售 | | 5 | nvshen | 18 | male | 技术 | +----+---------+------+--------+--------------+ 5 rows in set (0.00 sec)
#上述sql等同于 mysql> select employee.id,employee.name,employee.age,employee.sex,department.name from employee,department where employee.dep_id=department.id;
（3）外链接之左连接：优先显示左表全部记录
#以左表为准，即找出所有员工信息，当然包括没有部门的员工 #本质就是：在内连接的基础上增加左边有，右边没有的结果 mysql> select employee.id,employee.name,department.name as depart_name from employee left join department on employee.dep_id=department.id; +----+----------+--------------+ | id | name | depart_name | +----+----------+--------------+ | 1 | egon | 技术 | | 5 | nvshen | 技术 | | 2 | alex | 人力资源 | | 3 | wupeiqi | 人力资源 | | 4 | yuanhao | 销售 | | 6 | xiaomage | NULL | +----+----------+--------------+ 6 rows in set (0.00 sec)
（4） 外链接之右连接：优先显示右表全部记录
#以右表为准，即找出所有部门信息，包括没有员工的部门 #本质就是：在内连接的基础上增加右边有，左边没有的结果 mysql> select employee.id,employee.name,department.name as depart_name from employee right join department on employee.dep_id=department.id; +------+---------+--------------+ | id | name | depart_name | +------+---------+--------------+ | 1 | egon | 技术 | | 2 | alex | 人力资源 | | 3 | wupeiqi | 人力资源 | | 4 | yuanhao | 销售 | | 5 | nvshen | 技术 | | NULL | NULL | 运营 | +------+---------+--------------+ 6 rows in set (0.00 sec)
（5） 全外连接：显示左右两个表全部记录(了解)
#外连接：在内连接的基础上增加左边有右边没有的和右边有左边没有的结果
#注意：mysql不支持全外连接 full JOIN
#强调：mysql可以使用此种方式间接实现全外连接语法：select * from employee left join department on employee.dep_id = department.id
union all
select * from employee right join department on employee.dep_id = department.id;
mysql> select * from employee left join department on employee.dep_id = department.id union select * from employee right join department on employee.dep_id = department.id ; +------+----------+--------+------+--------+------+--------------+ | id | name | sex | age | dep_id | id | name | +------+----------+--------+------+--------+------+--------------+ | 1 | egon | male | 18 | 200 | 200 | 技术 | | 5 | nvshen | male | 18 | 200 | 200 | 技术 | | 2 | alex | female | 48 | 201 | 201 | 人力资源 | | 3 | wupeiqi | male | 38 | 201 | 201 | 人力资源 | | 4 | yuanhao | female | 28 | 202 | 202 | 销售 | | 6 | xiaomage | female | 18 | 204 | NULL | NULL | | NULL | NULL | NULL | NULL | NULL | 203 | 运营 | +------+----------+--------+------+--------+------+--------------+ 7 rows in set (0.01 sec) #注意 union与union all的区别：union会去掉相同的纪录
二、符合条件连接查询
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，即找出年龄大于25岁的员工以及员工所在的部门
select employee.name,department.name from employee inner join department on employee.dep_id = department.id where age > 25;
三、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：= 、 !=、> 、<等
（1）带in关键字的子查询
#查询平均年龄在25岁以上的部门名 select id,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 in (select id from department where name='技术'); #查看不足1人的部门名 select name from department where id not in (select dep_id from employee group by dep_id);
（2）带比较运算符的子查询
#比较运算符：=、!=、>、>=、<、<=、<> #查询大于所有人平均年龄的员工名与年龄 mysql> select name,age from employee where age > (select avg(age) from employee); +---------+------+ | name | age | +---------+------+ | alex | 48 | | wupeiqi | 38 | +---------+------+ #查询大于部门内平均年龄的员工名、年龄
思路：
（1）先对员工表(employee)中的人员分组(group by)，查询出dep_id以及平均年龄。
(2)将查出的结果作为临时表,再对根据临时表的dep_id和employee的dep_id作为筛选条件将employee表和临时表进行内连接。
(3)最后再将employee员工的年龄是大于平均年龄的员工名字和年龄筛选。
mysql> select t1.name,t1.age from employee as t1 inner join (select dep_id,avg(age) as avg_age from employee group by dep_id) as t2 on t1.dep_id = t2.dep_id where t1.age > t2.avg_age; +------+------+ | name | age | +------+------+ | alex | 48 |
(3)带EXISTS关键字的子查询
#EXISTS关字键字表示存在。在使用EXISTS关键字时，内层查询语句不返回查询的记录。而是返回一个真假值。True或False #当返回True时，外层查询语句将进行查询；当返回值为False时，外层查询语句不进行查询 #department表中存在dept_id=203，Ture mysql> select * from employee where exists (select id from department where id=200); +----+----------+--------+------+--------+ | id | name | sex | age | dep_id | +----+----------+--------+------+--------+ | 1 | egon | male | 18 | 200 | | 2 | alex | female | 48 | 201 | | 3 | wupeiqi | male | 38 | 201 | | 4 | yuanhao | female | 28 | 202 | | 5 | nvshen | male | 18 | 200 | | 6 | xiaomage | female | 18 | 204 | +----+----------+--------+------+--------+ #department表中存在dept_id=205，False mysql> select * from employee where exists (select id from department where id=204); Empty set (0.00 sec)
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