Question

我有两张桌子

表1

\n

AND

表2

      SNO RECEIVER USER1 USER2
      1   133       44   45
      2   144       66   77
      3   155       77   33

必填：从table1中删除table2中没有匹配条目的所有行

例如

     RECEIVER USER
     133      44
     133      45
     144      88
     144      55
     155      77

应该删除因为table2上没有144 66行

155 33

我试过但它不起作用 DELETE FROM TABLE1 LEFT JOIN TABLE2 B ON（A.RECEIVER = B.RECEIVER和A.USER1 = B.USER）和B.USER为空;

Answer 1

您也可以在DELETE语句中使用JOIN。以下语句连接（使用左连接！）两个表，从表2（WHERE ...）中选择没有相应记录的结果行，并从t1中删除相应的行：

DELETE t1
FROM t1 LEFT JOIN t2 ON (t1.RECEIVER = t2.RECEIVER AND t1.USER1 = t2.USER)
WHERE t2.RECEIVER IS NULL

请参阅manual以供参考，尤其是＆＃34;多表删除＆＃34;部分。

Answer 2

为什么需要JOIN来执行此操作，而不是使用SUB-QUERY

 DELETE FROM  table1
     WHERE receiver NOT IN ( SELECT DISTINCT (receiver) FROM table2 )
     AND user1 NOT IN ( SELECT DISTINCT (user) FROM table2 );

Answer 3

Delete * from Table1 
Where not exists (select 1 from Table2 where 
RECEIVER =  Table1.RECEIVER and (user = Table1.user1 or user = Table1.user2))

澄清不匹配的含义。您认为匹配什么？在我上面的回答中，我假设匹配意味着相同的接收器，user1或user2与Table2中的用户相同。如果不是这种情况，则可以修改查询。

Answer 4

像：

SELECT t1.sno FROM table1 t1 LEFT JOIN table2 t2 ON t1.receiver = t2.receiver and 2.user = t1.user1 WHERE t2.receiver IS NULL;

以上查询通过接收者列和用户列连接这两个表，并仅选择那些没有匹配记录的记录（left join，选择table1中的所有记录，而WHERE t2.receiver IS NULL将仅选择那些没有匹配记录的记录表2中的匹配记录

现在你可以做到

delete from table1 where sno IN (__PASTE_ABOVE_QUERY__);

关于连接的Mysql查询

4 个答案: