删除另一个表中不存在 id 的所有行答案

【问题标题】：Delete all rows which has no id existing in another table删除另一个表中不存在 id 的所有行
【发布时间】：2013-12-26 01:48:35
【问题描述】：

我想删除另一个表中没有现有外键的所有行示例：

table1
+----+-------+
|id  | data  |
+----+-------+
| 1  | hi    |
+----+-------+
| 2  | hi    |
+----+-------+
| 3  | hi    |
+----+-------+
| 4  | hi    |
+----+-------+
| 5  | hi    |
+----+-------+

table2
+----+-------+
|a_id| data  |
+----+-------+
| 1  | hi    |
+----+-------+
| 20 | hi    |
+----+-------+
| 3  | hi    |
+----+-------+
| 40 | hi    |
+----+-------+
| 5  | hi    |
+----+-------+

查询将删除 table2 上 id# 20 和 40 的行。

我需要这样做，以便我可以与 table1 和 table2 建立关系。

【问题讨论】：

Delete From Tab2 where ID not in (Select ID From Tab1)?

标签： mysql sql database-relations

【解决方案1】：

DELETE table2 
FROM   table2 
       LEFT JOIN table1 
              ON table2.a_id = table1.id 
WHERE  table1.id IS NULL

【讨论】：

【解决方案2】：

总结一下，删除多表有tree方式

NOT IN(SELECT ...) - @someone（他已经删除了他的答案）
```
Delete From Tab2 where ID not in (Select ID From Tab1)
```

LEFT JOIN - @eggyal

DELETE table2
FROM   table2 LEFT JOIN table1 ON table2.a_id = table1.id
WHERE  table1.id IS NULL

NOT EXISTS

DELETE
FROM  table2
WHERE NOT EXISTS (
  SELECT 1
  FROM table1
  WHERE table1.id = table2.a_id
)

根据What's the difference between NOT EXISTS vs. NOT IN vs. LEFT JOIN WHERE IS NULL?，不同的RDBMS表现不同。

【讨论】：