mysql 提示:Every derived table must have its own alias
2017-01-12 20:28
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Every derived table must have its own alias
这句话的中文意思是:每个派生出来的表都必须有一个自己的别名
问题的发生:一般出现在多表查询的时候,会出现此类错误。
因为,进行嵌套查询的时候子查询出来的的结果是作为一个派生表来进行上一级的查询的,所以子查询的结果必须要有一个别名
把MySQL语句改成:select count(*) from (select * from ……) as total;
问题就解决了,虽然只加了一个没有任何作用的别名total,但这个别名是必须的
示例SQL
select * from `user` a LEFT JOIN(select b.id from f_productinfo b where b.del_flag=0 and b.id=26) on a.USER_ID =b.id
错误截图:
解决办法:
正确sql:
select * from `user` a LEFT JOIN(select b.id from f_productinfo b where b.del_flag=0 and b.id=26) as c on a.USER_ID =c.id
这句话的中文意思是:每个派生出来的表都必须有一个自己的别名
问题的发生:一般出现在多表查询的时候,会出现此类错误。
因为,进行嵌套查询的时候子查询出来的的结果是作为一个派生表来进行上一级的查询的,所以子查询的结果必须要有一个别名
把MySQL语句改成:select count(*) from (select * from ……) as total;
问题就解决了,虽然只加了一个没有任何作用的别名total,但这个别名是必须的
示例SQL
select * from `user` a LEFT JOIN(select b.id from f_productinfo b where b.del_flag=0 and b.id=26) on a.USER_ID =b.id
错误截图:
解决办法:
正确sql:
select * from `user` a LEFT JOIN(select b.id from f_productinfo b where b.del_flag=0 and b.id=26) as c on a.USER_ID =c.id
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