SQL中使用update inner join和delete inner join

SQL中使用update inner join和delete inner join

Update

Update XXX set XXX where 这种写法大家肯定都知道，才发现update和delete居然支持inner join的update方式，这个在表间关联来做更新和删除操作非常有用.

列子:

Sql代码

update tb_User

set pass=''

from tb_User usr

inner join tb_Address addr on usr.nAddressFK = addr.nAddressID

where usr.id=123

update tb_User set pass='' from tb_User usr inner join tb_Address addr on usr.nAddressFK = addr.nAddressID where usr.id=123

update的格式是

update t1 set t1.name=’Liu’ from t1 inner join t2 on t1.id = t2.tid

MYSQL，ACCESS 写法如下:

Sql代码

UPDATE mem_world AS mw1 INNER JOIN mem_world AS mw2

ON mw1.parentid = mw2.wid

SET mw1.level = mw2.level

WHERE mw2.baseid = 107

AND mw2.parentid = 0

AND mw2.size > 1;

UPDATE mem_world AS mw1 INNER JOIN mem_world AS mw2 ON mw1.parentid = mw2.wid SET mw1.level = mw2.level WHERE mw2.baseid = 107 AND mw2.parentid = 0 AND mw2.size > 1;

on是表连接的筛选条件

就是说，表连接后，会产生一个类似于临时的视图这么一个东西

where是从这个临时的视图中筛选数据的

所以，你首先要搞清，你的所谓的2个条件属于哪一种

Delete

delete 语句也是类似

delete from t1 from t1 inner join t2 on t1.id = t2.tid

注意蓝色部分。

mysql:

Sql代码

DELETE mwb FROM mem_world_building AS mwb INNER JOIN mem_world AS mw

ON mwb.wid = mw.wid

where mw.type between 11 and 15

and baseid = 107

and mw.parentid <> 0

and mw.size > 1;

DELETE mwb FROM mem_world_building AS mwb INNER JOIN mem_world AS mw ON mwb.wid = mw.wid where mw.type between 11 and 15 and baseid = 107 and mw.parentid <> 0 and mw.size > 1;

下面是ORACLE的:

Sql代码

DELETE TABLE1 where KHID exists ( select KHID from table2 where FWDWID=8)

Sql代码

DELETE TABLE1 where exists ( select 1 from table2 where and table1.khid=table2.khid and FWDWID=8);