Question

我有3张桌子

employees     
-------------   
|idEmployee |  
|   name    |  

user-employee   
----------------                           
|  idUser      |-> MongoID      
|  idEmployee  |-> FK employees.idEmployee    

employees-supervisor
--------------------
|    idEmployee    |-> FK employees.idEmployee          
|    idUser        |-> FK user-employee.idUser

我想要下一张桌子

idEmployee | name   | idUser | nameSupervisor
    1      | John   |     A2 |     Jane

员工表包含公司的所有员工，用户员工拥有作为系统用户的所有员工，user-employee.idUser是mongo文档的用户，employees-supervisor具有匹配的列表idUser（mongoid），其id为employee（mysql）

我想要的是向所有员工展示他们各自的主管（如果有的话），现在我得到了idUser

SELECT employees.*, e1.idUser as idSupervisor, e2.name as supervisor
  FROM employees
  left JOIN user-employee e1 ON  e1.idEmployee = employees.idEmployee

显示

idEmployee | name   |  idUser | 
    1      | John   |  5887Ab |

我已经拥有主管的ID，现在我如何将该ID与用户 - 员工表上的名称相匹配？

Answer 1

如果密钥不匹配我们可以使用左连接

SELECT employees.*, e1.idUser as idSupervisor, e3.name as supervisor
FROM employees
left JOIN `user-employee` e1 ON  e1.idEmployee = employees.idEmployee 
left JOIN `employees-supervisor` e2 on on e2.idUser = e.idUser
left JOIN  employees e3 on e2.idEmployee = e3.idEmployee;

但如果表格中的关键字匹配，则应使用内连接

SELECT employees.*, e1.idUser as idSupervisor, e3.name as supervisor
FROM employees
INNER JOIN `user-employee` e1 ON  e1.idEmployee = employees.idEmployee 
INNER JOIN `employees-supervisor` e2 on on e2.idUser = e.idUser
INNER JOIN  employees e3 on e2.idEmployee = e3.idEmployee;

实施3表连接

1 个答案: