编程之路：MySql系列之多表查询

#重点：外链接语法

SELECT 字段列表
    FROM 表1 INNER|LEFT|RIGHT JOIN 表2
    ON 表1.字段 = 表2.字段;

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 | liwenzhou  | male   |   18 |    200 |  200 | 技术         |
|  5 | liwenzhou  | male   |   18 |    200 |  201 | 人力资源     |
|  5 | liwenzhou  | male   |   18 |    200 |  202 | 销售         |
|  5 | liwenzhou  | male   |   18 |    200 |  203 | 运营         |
|  6 | jingliyang | female |   18 |    204 |  200 | 技术         |
|  6 | jingliyang | female |   18 |    204 |  201 | 人力资源     |
|  6 | jingliyang | female |   18 |    204 |  202 | 销售         |
|  6 | jingliyang | female |   18 |    204 |  203 | 运营         |
+----+------------+--------+------+--------+------+--------------+

#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
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 | liwenzhou |   18 | male   | 技术         |
+----+-----------+------+--------+--------------+

#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;

#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 | liwenzhou  | 技术         |
|  2 | alex       | 人力资源     |
|  3 | wupeiqi    | 人力资源     |
|  4 | yuanhao    | 销售         |
|  6 | jingliyang | NULL         |
+----+------------+--------------+

#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 | liwenzhou | 技术         |
| NULL | NULL      | 运营         |
+------+-----------+--------------+

全外连接：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
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
;
#查看结果
+------+------------+--------+------+--------+------+--------------+
| id   | name       | sex    | age  | dep_id | id   | name         |
+------+------------+--------+------+--------+------+--------------+
|    1 | egon       | male   |   18 |    200 |  200 | 技术         |
|    5 | liwenzhou  | 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 | jingliyang | female |   18 |    204 | NULL | NULL         |
| NULL | NULL       | NULL   | NULL |   NULL |  203 | 运营         |
+------+------------+--------+------+--------+------+--------------+

#注意 union与union all的区别：union会去掉相同的纪录

#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;

#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

#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(子查询得到的是有人的部门id)
select name from department where id not in (select distinct dep_id from employee);

#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 emp where age > (select avg(age) from emp);
+---------+------+
| name | age |
+---------+------+
| alex | 48 |
| wupeiqi | 38 |
+---------+------+
rows in set (0.00 sec)


#查询大于部门内平均年龄的员工名、年龄
select t1.name,t1.age from emp t1
inner join 
(select dep_id,avg(age) avg_age from emp group by dep_id) t2
on t1.dep_id = t2.dep_id
where t1.age > t2.avg_age;

#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 | liwenzhou  | male   |   18 |    200 |
|  6 | jingliyang | 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)