MySQL从数据库中选择不在

如果在另一个数据库中不存在记录，我需要根据该数据库清理这些记录。
这很难解释，所以这里有一个例子：

Table Users
-----------
id
username
password

Table Articles
--------------
id
title
created_by
edited_by

---

created_by

和

deleted_by

包含用户ID。 我有3-4个表，它们的结构与文章表几乎相同，我想从表中删除那些在文章表中没有任何记录的用户。

---

我的意思是在

创建人

和

编辑人

表格中的

表格等文章中找不到ID的用户。

怎么做？

我首先尝试查看是否可以根据用户选择所有表的所有数据，但服务器无法执行查询：

SELECT * FROM `users` 
JOIN `articles` 
ON `articles`.`created_by` = `users`.`id`  
AND `articles`.`edited_by` = `users`.`id` 
JOIN `articles_two` 
ON `articles_two`.`created_by` = `users`.`id` 
AND `articles_two`.`edited_by` = `users`.`id` 
JOIN `articles_three` 
ON `articles_three`.`created_by` = `users`.`id` 
AND `articles_three`.`edited_by` = `users`.`id` 
JOIN `articles_four` 
ON `articles_four`.`created_by` = `users`.`id` 
AND `articles_four`.`edited_by` = `users`.`id` 
JOIN `articles_five` 
ON `articles_five`.`created_by` = `users`.`id` 
AND `articles_five`.`edited_by` = `users`.`id` 
JOIN `articles_six` 
ON `articles_six`.`created_by` = `users`.`id` 
AND `articles_six`.`edited_by` = `users`.`id`;

这应该行得通。它不是非常优雅，但我认为它很容易遵循：

DELETE FROM Users
WHERE ID NOT IN (
  SELECT Created_By FROM Articles
  UNION SELECT Edited_By FROM Articles
  UNION SELECT Created_By FROM Articles_Two
  UNION SELECT Edited_By FROM Articles_Two
  ...
  UNION SELECT Created_By FROM Articles_Six
  UNION SELECT Edited_By FROM Articles_Six
)

与任何大的“清理”查询一样，（a）首先复制一份表，（b）在键入
COMMIT
之前仔细检查，这应该是可行的。它不是非常优雅，但我认为它很容易遵循：

DELETE FROM Users
WHERE ID NOT IN (
  SELECT Created_By FROM Articles
  UNION SELECT Edited_By FROM Articles
  UNION SELECT Created_By FROM Articles_Two
  UNION SELECT Edited_By FROM Articles_Two
  ...
  UNION SELECT Created_By FROM Articles_Six
  UNION SELECT Edited_By FROM Articles_Six
)

与任何大型“清理”查询一样，（a）首先复制一份表，（b）在键入
COMMIT
之前仔细检查，我认为最干净的方法是
不在
选择
子句中：

select *
from users u
where u.id not in (select created_by from articles where created_by is not null) and
      u.id not in (select edited_by from articles where edited_by is not null) and
      u.id not in (select created_by from articles_two where created_by is not null) and
      u.id not in (select edited_by from articles_two where edited_by is not null) and
      u.id not in (select created_by from articles_three where created_by is not null) and
      u.id not in (select edited_by from articles_three where edited_by is not null) and
      u.id not in (select created_by from articles_four where created_by is not null) and
      u.id not in (select edited_by from articles_four where edited_by is not null)

在由
创建的和由
编辑的各个列上建立索引，应该有助于提高性能。
我认为最干净的方法是在
选择
子句中的
不在
中：

select *
from users u
where u.id not in (select created_by from articles where created_by is not null) and
      u.id not in (select edited_by from articles where edited_by is not null) and
      u.id not in (select created_by from articles_two where created_by is not null) and
      u.id not in (select edited_by from articles_two where edited_by is not null) and
      u.id not in (select created_by from articles_three where created_by is not null) and
      u.id not in (select edited_by from articles_three where edited_by is not null) and
      u.id not in (select created_by from articles_four where created_by is not null) and
      u.id not in (select edited_by from articles_four where edited_by is not null)

性能应该通过在由
创建的
和由
编辑的
列上建立索引来提高。
在MySQL中，
左外连接。。。如果null
往往比
中的

或存在的性能更好（如果有适当的索引），那么以下内容应该值得尝试：

SELECT u.* FROM `users` u
LEFT JOIN `articles` ac1 ON ac1.`created_by` = u.`id`  
LEFT JOIN `articles_two` ac2 ON ac2.`created_by` = u.`id` 
LEFT JOIN `articles_three` ac3 ON ac3.`created_by` = u.`id` 
LEFT JOIN `articles_four` ac4 ON ac4.`created_by` = u.`id` 
LEFT JOIN `articles_five` ac5 ON ac5.`created_by` = u.`id` 
LEFT JOIN `articles_six` ac6 ON ac6.`created_by` = u.`id` 
LEFT JOIN `articles` ae1 ON ae1.`edited_by` = u.`id` 
LEFT JOIN `articles_two` ae2 ON ae2.`edited_by` = u.`id` 
LEFT JOIN `articles_three` ae3 ON ae3.`edited_by` = u.`id` 
LEFT JOIN `articles_four` ae4 ON ae4.`edited_by` = u.`id` 
LEFT JOIN `articles_five` ae5 ON ae5.`edited_by` = u.`id` 
LEFT JOIN `articles_six` ae6 ON ae6.`edited_by` = u.`id`
WHERE COALESCE(ac1.`created_by`,ac2.`created_by`,ac3.`created_by`,
               ac4.`created_by`,ac5.`created_by`,ac6.`created_by`,
               ae1.`edited_by`, ae2.`edited_by`, ae3.`edited_by`, 
               ae4.`edited_by`, ae5.`edited_by`, ae6.`edited_by`)
IS NULL;

---

在MySQL中，

左外连接。。。如果null

往往比中的或存在的性能更好（如果有适当的索引），那么以下内容应该值得尝试：

SELECT u.* FROM `users` u
LEFT JOIN `articles` ac1 ON ac1.`created_by` = u.`id`  
LEFT JOIN `articles_two` ac2 ON ac2.`created_by` = u.`id` 
LEFT JOIN `articles_three` ac3 ON ac3.`created_by` = u.`id` 
LEFT JOIN `articles_four` ac4 ON ac4.`created_by` = u.`id` 
LEFT JOIN `articles_five` ac5 ON ac5.`created_by` = u.`id` 
LEFT JOIN `articles_six` ac6 ON ac6.`created_by` = u.`id` 
LEFT JOIN `articles` ae1 ON ae1.`edited_by` = u.`id` 
LEFT JOIN `articles_two` ae2 ON ae2.`edited_by` = u.`id` 
LEFT JOIN `articles_three` ae3 ON ae3.`edited_by` = u.`id` 
LEFT JOIN `articles_four` ae4 ON ae4.`edited_by` = u.`id` 
LEFT JOIN `articles_five` ae5 ON ae5.`edited_by` = u.`id` 
LEFT JOIN `articles_six` ae6 ON ae6.`edited_by` = u.`id`
WHERE COALESCE(ac1.`created_by`,ac2.`created_by`,ac3.`created_by`,
               ac4.`created_by`,ac5.`created_by`,ac6.`created_by`,
               ae1.`edited_by`, ae2.`edited_by`, ae3.`edited_by`, 
               ae4.`edited_by`, ae5.`edited_by`, ae6.`edited_by`)
IS NULL;

---

创建者和编辑器应该始终具有相同的

id

？只要您看到名为某物\u 1、某物\u 2等的表。。。警钟应该开始响了。您可能需要修改您的方案设计。短期内，您应该使用UNION将这些表“合并”（其中的相关列）到一个表中。

created_by

和

edited_by

可能会有所不同，这会让事情变得更困难。数据库设计真的很糟糕，但我无法进行更改：（创建者和编辑器应该始终具有相同的

id

？每当您看到名为\u 1、\u 2等的表时，就会响起警钟。您可能需要修改架构设计。短期内，您应该使用UNION来“合并”（中的相关列）将这些表合并为一个。

由

创建的和由编辑的可能不同，这使得事情变得更难。数据库设计非常糟糕，但我无法进行更改：(