SQL删除基于另一个表的行

Question

这可能很容易,但是周一早上.我有两张桌子:

表格1:

Field        | Type             | Null | Key | Default | Extra
id           | int(32) unsigned | NO   | PRI | NULL    | auto_increment
group        | int(32)          | NO   |     | 0       |                

表2:

Field     | Type             | Null | Key | Default | Extra
group     | int(32)          | NO   |     | 0       | 

忽略其他字段...我想要一个SQL DELETE语句,它将删除Table1中存在等于Table1.group的Table2.group的所有行.因此,如果Table1的一行具有group = 69,那么当且仅当Table2中存在group = 69的行时才应该删除该行.

感谢您的任何帮助.

Answer 1

我想这就是你想要的:

DELETE FROM `table1`
WHERE `group` in (SELECT DISTINCT `group` FROM `table2`)

Answer 2

我觉得这种方式更快:

DELETE FROM t1 USING table1 t1 INNER JOIN table2 t2 ON ( t1.group = t2.group );

Answer 3

好的解决方案就是编写SQL,就像你自己说的那样:

DELETE FROM Table1
WHERE
  EXISTS(SELECT 1 FROM Table2 WHERE Table2.Group = Table1.Group)

此致,Arno Brinkman

Answer 4

像这样的东西

delete from table1 where group in (select group from table2)