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  • python3 mysql-多表查询

    1、准备工作

      准备两张表,部门表和员工表:

    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)
    View Code

      注意:部门表中的id和员工表中的dep_id并不是一一对应的。

    2、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,多表连接查询。

      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)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(关于笛卡尔积的含义,大家百度自行补脑)。

    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         |
    View Code

      其他的情况就是在笛卡尔积的情况下添加约束条件,来得到自己想要的东西。

      (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;
    +----+---------+------+--------+--------------+
    | 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;
    View Code

      (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)
    View Code

      (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)
    View Code

      (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
    View Code

    3、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;

    4、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);
    View Code

      (2)带比较符运算的子查询

    #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 |
    View Code

      (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)
    View Code

      

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  • 原文地址:https://www.cnblogs.com/490144243msq/p/11186057.html
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