Sql 从两个表中获取用户，其中id'；它们是不一样的

我遇到了一个针对RESTAPI的漂亮查询。我想听听你的建议

现在我有两张桌子：

公司用户：存储注册用户的主要信息，如： 身份证、姓名、电子邮件、电话号码

匹配用户：存储以下信息： 活动用户标识，匹配用户标识，其中活动用户标识是授权用户标识，匹配用户标识是授权用户以前遇到的任何其他用户标识 问题是：如何执行查询，以便能够获取所有其他用户（授权用户ID除外，以及授权用户以前会遇到的用户）

例如：

表：公司用户

---

ID |姓名|电子邮件

---

1 |杰克|jake@gmail.com

---

2 |简|jane@gmail.com

---

3 |雅各布|jacob@gmail.com

---

4 |六月|june@gmail.com

及

表格：匹配表格

ID |活动用户|匹配用户| ID

---

1 | 1 | 2

---

如何为id为3和4的公司\u用户中的仅授权用户执行？

获取表2的所有行：

$matching_rows = MatchingTable::all();
$active_users = $matching_rows->pluck('active_user_id')->toArray();
$matching_users = $matching_rows->pluck('active_user_id')->toArray();

将它们合并在一起：

$auth_users = array_merge($active_users, $matching_users);

使用whereNotIn：

$unauth_users = Users::whereNotIn('id', $auth_users)->get()`

---

获取表2的所有行：

$matching_rows = MatchingTable::all();
$active_users = $matching_rows->pluck('active_user_id')->toArray();
$matching_users = $matching_rows->pluck('active_user_id')->toArray();

将它们合并在一起：

$auth_users = array_merge($active_users, $matching_users);

使用whereNotIn：

$unauth_users = Users::whereNotIn('id', $auth_users)->get()`

请试试这个

如果除活动用户ID之外，则

Select * from company_users
Where Id NOT IN ( SELECT Distinct ACTIVE_USER_ID  FROM matching_table)

Select * from company_users
Where Id NOT IN ( SELECT Distinct MATCHING_USER_ID FROM matching_table)

Select * from company_users
Where Id NOT IN ( SELECT Distinct MATCHING_USER_ID  FROM matching_table)
AND Id NOT IN ( SELECT Distinct ACTIVE_USER_ID  FROM matching_table)

如果不匹配用户ID，则

Select * from company_users
Where Id NOT IN ( SELECT Distinct ACTIVE_USER_ID  FROM matching_table)

Select * from company_users
Where Id NOT IN ( SELECT Distinct MATCHING_USER_ID FROM matching_table)

Select * from company_users
Where Id NOT IN ( SELECT Distinct MATCHING_USER_ID  FROM matching_table)
AND Id NOT IN ( SELECT Distinct ACTIVE_USER_ID  FROM matching_table)

如果除活动用户ID和匹配用户ID之外，则

Select * from company_users
Where Id NOT IN ( SELECT Distinct ACTIVE_USER_ID  FROM matching_table)

Select * from company_users
Where Id NOT IN ( SELECT Distinct MATCHING_USER_ID FROM matching_table)

Select * from company_users
Where Id NOT IN ( SELECT Distinct MATCHING_USER_ID  FROM matching_table)
AND Id NOT IN ( SELECT Distinct ACTIVE_USER_ID  FROM matching_table)

请试试这个

如果除活动用户ID之外，则

Select * from company_users
Where Id NOT IN ( SELECT Distinct ACTIVE_USER_ID  FROM matching_table)

Select * from company_users
Where Id NOT IN ( SELECT Distinct MATCHING_USER_ID FROM matching_table)

Select * from company_users
Where Id NOT IN ( SELECT Distinct MATCHING_USER_ID  FROM matching_table)
AND Id NOT IN ( SELECT Distinct ACTIVE_USER_ID  FROM matching_table)

如果不匹配用户ID，则

Select * from company_users
Where Id NOT IN ( SELECT Distinct ACTIVE_USER_ID  FROM matching_table)

Select * from company_users
Where Id NOT IN ( SELECT Distinct MATCHING_USER_ID FROM matching_table)

Select * from company_users
Where Id NOT IN ( SELECT Distinct MATCHING_USER_ID  FROM matching_table)
AND Id NOT IN ( SELECT Distinct ACTIVE_USER_ID  FROM matching_table)

如果除活动用户ID和匹配用户ID之外，则

Select * from company_users
Where Id NOT IN ( SELECT Distinct ACTIVE_USER_ID  FROM matching_table)

Select * from company_users
Where Id NOT IN ( SELECT Distinct MATCHING_USER_ID FROM matching_table)

Select * from company_users
Where Id NOT IN ( SELECT Distinct MATCHING_USER_ID  FROM matching_table)
AND Id NOT IN ( SELECT Distinct ACTIVE_USER_ID  FROM matching_table)