Question

我有两个表（所有者和员工）和一个映射表（owner_employee）

我需要一个select语句，通过 employee_id 获取所有所有者员工，而不使用内部选择

示例：

SELECT * FROM employee e
[...Joins here...]
WHERE e.id = 1

应该返回

owner | employee
1     |   1
1     |   2

所有者表

id | title
----------
1  |  a
2  |  b
3  |  c

员工表

id | title
----------
1  |  a
2  |  a
3  |  b
4  |  c

Owner_Employee

id_owner | id_employee
----------------------
1        |    1
1        |    2
2        |    3
3        |    4

这里内部选择的工作实现（可能正在工作，因为我从原始示例重写了它）

SELECT *
 FROM owner_employee oe
 JOIN owner o  ON o.id = oe.id_owner                                                 
 JOIN employee e ON e.id = oe.id_employee
 WHERE o.id in 
 (SELECT id_owner 
  FROM owner_employee oe
  WHERE oe.id_employee in (1))

Answer 1

也许是这样的：

 SELECT DISTINCT oe.*,o.*,e.*
 FROM owner_employee AS z 
 JOIN owner_employee oe on oe.id_owner = z.id_owner
 JOIN owner o  ON o.id = oe.id_owner
 JOIN employee e ON e.id = oe.id_employee
 WHERE z.id_employee in (1);

Answer 2

您提供的查询代码（不会返回您的文本要求的内容）与

相同

SELECT *
FROM owner_employee oe
JOIN owner o  ON o.id = oe.id_owner                                                 
JOIN employee e ON e.id = oe.id_employee
WHERE oe.id_employee = 1

但是您的文字要求的是＆＃34;所有所有者员工都是employee_id＆＃34;显然没有意义。由于您的数据，一个所有者可以拥有多个id_employee。

如果你想要＆＃34;所有id_owners及其id_employees＆＃34;：

SELECT * FROM owner_employee

如果您想要＆＃34;所有所有者的id_employees＆＃34;：

SELECT id_employee FROM owner_employee

或许你想要＆＃34;所有的所有者，每个人都是他们的员工ID＆＃34;？

按子表id连接表而不选择内部表

2 个答案: