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这个sql语句该如何修改下

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Release： 2016-06-13 13:47:53
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这个sql语句该怎么修改下？

SQL code
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-->SELECT si.user_id FROM ecs_store_info AS si WHERE si.store_name like (SELECT wl.store_name FROM ecs_wpaqq_log AS wl WHERE wl.lngShopId=0)
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就是如何在like关键后使用select语句

------解决方案--------------------
LZ的想法我大致明白，但是like是模糊匹配一串字符。LZ如果是想在很多个字符里面，随便模糊匹配任意一个，只能用in或者通过PHP遍历结果然后重新sql，这个目前没见过如何在一条sql里面实现的，不过应该有大神能做到吧，这我就不清楚了。
------解决方案--------------------
用下面这句，我在Oracle里面测试是可以的：
select * from users u where u.name like '%'||(select name from users where id=1)||'%';
------解决方案--------------------
LS，如果子查询中只能取出唯一一条数据，那么LZ的问题就不是问题了。

问题的关键就在于他LIKE中匹配的是个结果集...我刚才查了下，mysql中的函数也没有能模糊匹配一串字符

的，LZ你放弃吧，最好换写法，因为就算用sql语句实现了，效率也会出奇的底下。
------解决方案--------------------
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SQL code
SELECT si.user_id FROM ecs_store_info AS si WHERE si.store_name like (SELECT wl.store_name FROM ecs_wpaqq_log AS wl WHERE wl.lngShopId=0)
就是如何在like关键后使用select语句
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from like select store where
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