python--MySQL多表查询

#建表
#部门表
create table department(
id int,
name varchar(20) 
);

#员工表，之前我们学过foreign key，强行加上约束关联，但是我下面这个表并没有直接加foreign key，这两个表我只是让它们在逻辑意义上有关系，并没有加foreign key来强制两表建立关系，为什么要这样搞，是有些效果要给大家演示一下
#所以，这两个表是不是先建立哪个表都行啊，如果有foreign key的话，是不是就需要注意表建立的顺序了。那我们来建表。
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
);

#给两个表插入一些数据
insert into department values
(200,‘技术‘),
(201,‘人力资源‘),
(202,‘销售‘),
(203,‘运营‘); #注意这一条数据，在下面的员工表里面没有对应这个部门的数据

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),
(‘liwenzhou‘,‘male‘,18,200),
(‘jingliyang‘,‘female‘,18,204) #注意这条数据的dep_id字段的值，这个204，在上面的部门表里面也没有对应的部门id。所以两者都含有一条双方没有涉及到的数据，这都是为了演示一下效果设计的昂
;


#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 | |
+-------+-------------+------+-----+---------+-------+

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 | |
+--------+-----------------------+------+-----+---------+----------------+

mysql> select * from department;
+------+--------------+
| id | name |
+------+--------------+
| 200 | 技术 |
| 201 | 人力资源 |
| 202 | 销售 |
| 203 | 运营 |
+------+--------------+

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 | liwenzhou | male | 18 | 200 |
| 6 | jingliyang | female | 18 | 204 |
+----+------------+--------+------+--------+

# 重点: 外链接语法

select 字段列表  from 表1  inner 或 left 或 right  join 表2  on 表1.字段  =  表2.字段;
　　# 优先级: on在join前, join在where前, 改名的as在on之前

mysql> select * from department,employee; #表用逗号分隔，看我查询时表的顺序，先department后employee，所以你看结果表的这些字段，是不是就是我们两个表字段并且哪个表在前面，哪个表的字段就在前面
+------+--------------+----+------------+--------+------+--------+
| id   | name         | id | name       | sex    | age  | dep_id |
+------+--------------+----+------------+--------+------+--------+
|  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         |  1 | egon       | male   |   18 |    200 |
|  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         |  2 | alex       | female |   48 |    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         |  3 | wupeiqi    | male   |   38 |    201 |
|  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         |  4 | yuanhao    | female |   28 |    202 |
|  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         |  5 | liwenzhou  | male   |   18 |    200 |
|  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         |  6 | jingliyang | female |   18 |    204 |
+------+--------------+----+------------+--------+------+--------+
rows in set (0.12 sec)

我们让employee表在前面看看结果，注意看结果表的字段
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         |
+----+------------+--------+------+--------+------+--------------+
rows in set (0.00 sec)

关于笛卡儿积：从上面的数据可以看出来,笛卡尔积就是表1 的每一行分别对应表2 的每一行.

　　因为我们要进行连表查询，那么mysql并不知道你想要如何连接两个表的关系进行查询，那么mysql会将你两个表数据的所有组合关系都给你拼接成一条数据来显示，这样你就可以想查哪个关联关系的数据就查哪个了，如果还是不太理解看一下下面的图：

#我们要找的数据就是员工表里面dep_id字段的值和部门表里面id字段的值能对应上的那些数据啊，所以你看下面的写法：
mysql> select * from employee,department where employee.dep_id=department.id;
+----+-----------+--------+------+--------+------+--------------+
| id | name      | sex    | age  | dep_id | id   | name         |
+----+-----------+--------+------+--------+------+--------------+
|  1 | egon      | 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         |
|  5 | liwenzhou | male   |   18 |    200 |  200 | 技术         |
+----+-----------+--------+------+--------+------+--------------+
rows in set (0.14 sec)
拿到了我们想要的结果。

但是你看，我们左表employee表中的dep_id为204的那个数据没有了，右表department表的id为203的数据没有了，因为我们现在要的就是两表能对应上的数据一起查出来，那个204和203双方对应不上。

#再看一个需求，我要查出技术部的员工的名字
mysql> select name from employee,department where employee.dep_id=department.id and department.name=‘技术‘;
ERROR 1052 (23000): Column ‘name‘ in field list is ambiguous
#上面直接就报错了，因为select后面直接写的name，在两个表合并起来的表中，是有两个name字段的，直接写name是不行的，要加上表名，再看：
mysql> select employee.name from employee,department where employee.dep_id=department.id and department.name=‘技术‘;
+-----------+
| name      |
+-----------+
| egon      |
| liwenzhou |
+-----------+
rows in set (0.09 sec)
结果就没问题了

#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  #注意语法：
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，因为union all是left join 和right join合并，所以有重复的记录，通过union就将重复的记录去重了。

#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;

#我们之前的做法是：先连表
mysql> select * from employee inner join department on employee.dep_id = department.id; 
+----+-----------+--------+------+--------+------+--------------+
| id | name      | sex    | age  | dep_id | id   | name         |
+----+-----------+--------+------+--------+------+--------------+
|  1 | egon      | 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         |
|  5 | liwenzhou | male   |   18 |    200 |  200 | 技术         |
+----+-----------+--------+------+--------+------+--------------+
rows in set (0.10 sec)

#然后根据连表的结果进行where过滤，将select*改为select employee.name

mysql> select employee.name from employee inner join department on employee.dep_id = department.id where department.name=‘技术‘;
  +-----------+
  | name      |
  +-----------+
  | egon      |
  | liwenzhou |
  +-----------+
rows in set (0.09 sec)

#首先从部门表里面找到技术部门对应的id
mysql> select id from department where name=‘技术‘;
+------+
| id   |
+------+
|  200 |
+------+
row in set (0.00 sec)

#那我们把上面的查询结果用括号括起来，它就表示一条id=200的数据，然后我们通过员工表来查询dep_id=这条数据作为条件来查询员工的name
mysql> select name from employee where dep_id = (select id from department where name=‘技术‘);
+-----------+
| name      |
+-----------+
| egon      |
| liwenzhou |
+-----------+
rows in set (0.00 sec)
上面这些就是子查询的一个思路，解决一个问题，再解决另外一个问题，你子查询里面可不可以是多个表的查询结果，当然可以，然后再通过这个结果作为依据来进行过滤，然后我们学一下子查询里面其他的内容，往下学。

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

#查询员工平均年龄在25岁以上的部门名，可以用连表，也可以用子查询，我们用子查询来搞一下
select id,name from department
    where id in 
        (select dep_id from employee group by dep_id having avg(age) > 25);
#连表来搞一下上面这个需求
select department.name from department inner join employee on department.id=employee.dep_id 
    group by department.name 
    having avg(age)>25;
总结：子查询的思路和解决问题一样，先解决一个然后拿着这个的结果再去解决另外一个问题，连表的思路是先将两个表关联在一起，然后在进行group by啊过滤啊等等操作，两者的思路是不一样的

#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)