我有一个包含 4 个表的数据库，我想加入它们，但我得到了错误的结果，而且我不明白为什么
这是我运行的 mysql 查询
SELECT co_id, u_name, c_name, e_status
FROM users
INNER JOIN company_owners ON u_id = co_userFk
INNER JOIN company ON co_companyFk = c_id
INNER JOIN employees ON u_id = userFk WHERE c_id = "1"
这是我得到的结果
+------------+-------------+-----------+------------------+
| co_id | u_name | c_name | e_status |
+------------+-------------+-----------+------------------+
| 1 | Simon | SimonCorp | Chef |
| 1 | Simon | SimonCorp | Chef |
+------------+-------------+-----------+------------------+
这就是我想要的结果
+------------+-------------+-----------+------------------+
| co_id | u_name | c_name | e_status |
+------------+-------------+-----------+------------------+
| 1 | Simon | SimonCorp | Chef |
+------------+-------------+-----------+------------------+
这是我的数据库结构
Users
+------------+-------------+----------+------------------+
| u_id | u_name | u_address| u_email |
+------------+-------------+----------+------------------+
| 1 | Simon | street 1 | Simon@gamil.com |
| 2 | Andrew | street 2 | Andrew@gmail.com |
| 3 | Tom | street 3 | Tom@gmail.com |
| 4 | Charlie | street 4 | charlie@gmail.com|
+------------+-------------+----------+------------------+
Company
+------------+-------------+----------+--------------------------+
| c_id | c_name | c_address| c_email |
+------------+---------------+----------+------------------------+
| 1 | SimonCorp | street 1 | simoncorp@gamil.com |
| 2 | SimonotherCorp| street 2 |simonothercorp@gmail.com|
+------------+---------------+----------+------------------------+
Employees
+------------+-------------+----------+----------------------+
| e_id | companyFk | userFk | e_status |
+------------+-------------+----------+----------------------+
| 1 | 1 | 1 | Chef |
| 2 | 2 | 2 | employee |
| 3 | 2 | 1 | employee |
| 4 | 1 | 3 | employee |
+------------+-------------+----------+----------------------+
Simon is chef of first company and the employee in the second
Company_Owners
+------------+--------------+-------------+
| co_id | co_companyFk | co_userFk |
+------------+--------------+-------------+
| 1 | 1 | 1 |
| 2 | 2 | 1 |
+------------+--------------+-------------+
Simon owns both companies
最佳答案
这是因为您的表Employees 仅考虑u_id = userFk 作为连接条件。
表“员工”有 2 行链接到该员工(西蒙):
Employees
+------------+-------------+----------+----------------------+
| e_id | companyFk | userFk | e_status |
+------------+-------------+----------+----------------------+
| 1 | 1 | 1 | Chef |
| 2 | 2 | 2 | employee |
| 3 | 2 | 1 | employee |
| 4 | 1 | 3 | employee |
+------------+-------------+----------+----------------------+
因此，对于每条生产线，将与作为公司所有者的 Simon 一起规划一个结果，考虑到该联合。
要解决此问题，您可以限制连接以也考虑 companyFK。
它会是这样的:
SELECT co_id, u_name, c_name, e_status
FROM users
INNER JOIN company_owners ON u_id = co_userFk
INNER JOIN company ON co_companyFk = c_id
INNER JOIN employees ON (u_id = userFk and c_id = companyFk) WHERE c_id = "1"
关于mysql - mysql新手，使用内连接得到错误的结果，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/42681792/