SQL删除行，其中一个值与DIFFERENT列中的值匹配

Question

如果特定x值与同一表中任何行中的y值匹配，我想删除一行。

前：

| x | y |  
| 4 | 2 |  
| 2 | 6 |  
| 8 | 1 |  
| 3 | 1 |  
| 7 | 8 |  
| 9 | 5 |

会变成：

| x | y |    
| 4 | 2 |  
| 3 | 1 |  
| 7 | 8 |  
| 9 | 5 |

Answer 1

使用EXISTS

Delete from 
yourtable where exists (select 1 from tab b where b.y =yourtable.x) 

Answer 2

如果您的数据库允许，可以使用自联接：

DELETE FROM foo AS xside
LEFT JOIN foo AS yside ON xside.y = yside.x

Answer 3

Delete from tab where x in ( select y from tab)

替代版本以计算y列中的空值。

Delete from tab t where exists ( select 1 from tab ta where ta.y = t.x)