本节重点：
- 多表连接查询
- 符合条件连接查询
- 子查询
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 ('tom','male',18,200), ('mike','female',48,201), ('jack','male',38,201), ('lucy','female',28,202), ('lili','male',18,200), ('alice','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 | tom | male | 18 | 200 | | 2 | mike | female | 48 | 201 | | 3 | jack | male | 38 | 201 | | 4 | lucy | female | 28 | 202 | | 5 | lili | male | 18 | 200 | | 6 | alice | 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、多表连接查询
两张表的准备工作已完成，比如现在我要查询的员工信息以及该员工所在的部门。从该题中，我们看出既要查员工又要查该员工的部门，肯定要将两张表进行连接查询。
重点：外链接语法
语法：
SELECT 字段列表
FROM 表1 INNER|LEFT|RIGHT JOIN 表2
ON 表1.字段 = 表2.字段;
1.1、交叉连接
不适用任何匹配条件。生成笛卡尔积（简单的说就是两个集合相乘的结果）。
mysql> select * from employee, department; +----+-------+--------+------+--------+------+--------------+ | id | name | sex | age | dep_id | id | name | +----+-------+--------+------+--------+------+--------------+ | 1 | tom | male | 18 | 200 | 200 | 技术 | | 1 | tom | male | 18 | 200 | 201 | 人力资源 | | 1 | tom | male | 18 | 200 | 202 | 销售 | | 1 | tom | male | 18 | 200 | 203 | 运营 | | 2 | mike | female | 48 | 201 | 200 | 技术 | | 2 | mike | female | 48 | 201 | 201 | 人力资源 | | 2 | mike | female | 48 | 201 | 202 | 销售 | | 2 | mike | female | 48 | 201 | 203 | 运营 | | 3 | jack | male | 38 | 201 | 200 | 技术 | | 3 | jack | male | 38 | 201 | 201 | 人力资源 | | 3 | jack | male | 38 | 201 | 202 | 销售 | | 3 | jack | male | 38 | 201 | 203 | 运营 | | 4 | lucy | female | 28 | 202 | 200 | 技术 | | 4 | lucy | female | 28 | 202 | 201 | 人力资源 | | 4 | lucy | female | 28 | 202 | 202 | 销售 | | 4 | lucy | female | 28 | 202 | 203 | 运营 | | 5 | lili | male | 18 | 200 | 200 | 技术 | | 5 | lili | male | 18 | 200 | 201 | 人力资源 | | 5 | lili | male | 18 | 200 | 202 | 销售 | | 5 | lili | male | 18 | 200 | 203 | 运营 | | 6 | alice | female | 18 | 204 | 200 | 技术 | | 6 | alice | female | 18 | 204 | 201 | 人力资源 | | 6 | alice | female | 18 | 204 | 202 | 销售 | | 6 | alice | female | 18 | 204 | 203 | 运营 | +----+-------+--------+------+--------+------+--------------+ 24 rows in set (0.06 sec)
1.2、内连接：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; 或 mysql> select * from employee inner join department on employee.dep_id = department.id; +----+------+--------+------+--------+------+--------------+ | id | name | sex | age | dep_id | id | name | +----+------+--------+------+--------+------+--------------+ | 1 | tom | male | 18 | 200 | 200 | 技术 | | 2 | mike | female | 48 | 201 | 201 | 人力资源 | | 3 | jack | male | 38 | 201 | 201 | 人力资源 | | 4 | lucy | female | 28 | 202 | 202 | 销售 | | 5 | lili | male | 18 | 200 | 200 | 技术 | +----+------+--------+------+--------+------+--------------+ 5 rows in set (0.05 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;
1.3、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 * from employee left join department on employee.dep_id = department.id; +----+-------+--------+------+--------+------+--------------+ | id | name | sex | age | dep_id | id | name | +----+-------+--------+------+--------+------+--------------+ | 1 | tom | male | 18 | 200 | 200 | 技术 | | 5 | lili | male | 18 | 200 | 200 | 技术 | | 2 | mike | female | 48 | 201 | 201 | 人力资源 | | 3 | jack | male | 38 | 201 | 201 | 人力资源 | | 4 | lucy | female | 28 | 202 | 202 | 销售 | | 6 | alice | female | 18 | 204 | NULL | NULL | +----+-------+--------+------+--------+------+--------------+ 6 rows in set (0.00 sec)
1.4、外链接之右连接：优先显示右表全部记录
mysql> select * from employee right join department on employee.dep_id = department.id; +------+------+--------+------+--------+------+--------------+ | id | name | sex | age | dep_id | id | name | +------+------+--------+------+--------+------+--------------+ | 1 | tom | male | 18 | 200 | 200 | 技术 | | 2 | mike | female | 48 | 201 | 201 | 人力资源 | | 3 | jack | male | 38 | 201 | 201 | 人力资源 | | 4 | lucy | female | 28 | 202 | 202 | 销售 | | 5 | lili | male | 18 | 200 | 200 | 技术 | | NULL | NULL | NULL | NULL | NULL | 203 | 运营 | +------+------+--------+------+--------+------+--------------+ 6 rows in set (0.00 sec)
1.5、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和右边都有的结果
#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
mysql> 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; +------+-------+--------+------+--------+------+--------------+ | id | name | sex | age | dep_id | id | name | +------+-------+--------+------+--------+------+--------------+ | 1 | tom | male | 18 | 200 | 200 | 技术 | | 5 | lili | male | 18 | 200 | 200 | 技术 | | 2 | mike | female | 48 | 201 | 201 | 人力资源 | | 3 | jack | male | 38 | 201 | 201 | 人力资源 | | 4 | lucy | female | 28 | 202 | 202 | 销售 | | 6 | alice | female | 18 | 204 | NULL | NULL | | 1 | tom | male | 18 | 200 | 200 | 技术 | | 2 | mike | female | 48 | 201 | 201 | 人力资源 | | 3 | jack | male | 38 | 201 | 201 | 人力资源 | | 4 | lucy | female | 28 | 202 | 202 | 销售 | | 5 | lili | male | 18 | 200 | 200 | 技术 | | NULL | NULL | NULL | NULL | NULL | 203 | 运营 | +------+-------+--------+------+--------+------+--------------+ 12 rows in set (0.00 sec) mysql> 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 | tom | male | 18 | 200 | 200 | 技术 | | 5 | lili | male | 18 | 200 | 200 | 技术 | | 2 | mike | female | 48 | 201 | 201 | 人力资源 | | 3 | jack | male | 38 | 201 | 201 | 人力资源 | | 4 | lucy | female | 28 | 202 | 202 | 销售 | | 6 | alice | female | 18 | 204 | NULL | NULL | | NULL | NULL | NULL | NULL | NULL | 203 | 运营 | +------+-------+--------+------+--------+------+--------------+ 7 rows in set (0.06 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
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;
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 employee.id,employee.name,employee.age,department.name from employee,department
where employee.dep_id = department.id
and age > 25
order by age asc;
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
3.1、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); +------+--------------+ | id | name | +------+--------------+ | 201 | 人力资源 | | 202 | 销售 | +------+--------------+ # 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'); +------+ | name | +------+ | tom | | lili | +------+ #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); 或 mysql> select name from department where id not in (select dep_id from employee); +--------+ | name | +--------+ | 运营 | +--------+ 1 row in set (0.00 sec)
3.2、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 | +------+------+ | mike | 48 | | jack | 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 name,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 | +------+------+ | mike | 48 | +------+------+ 1 row in set (0.07 sec)
3.3、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=200，True mysql> select * from employee where exists (select id from department where id=200); +----+-------+--------+------+--------+ | id | name | sex | age | dep_id | +----+-------+--------+------+--------+ | 1 | tom | male | 18 | 200 | | 2 | mike | female | 48 | 201 | | 3 | jack | male | 38 | 201 | | 4 | lucy | female | 28 | 202 | | 5 | lili | male | 18 | 200 | | 6 | alice | female | 18 | 204 | +----+-------+--------+------+--------+ #department表中不存在dept_id=204，False mysql> select * from employee where exists (select id from department where id=204); Empty set (0.00 sec)
