在使用MYSQL进行嵌套查询时出现:Every derived table must have its own alias错误解决方法

Posted 2021-10-04 浪子尘晨

在做多表查询，或者查询的时候产生新的表的时候会出现这个错误：Every derived table must have its own alias（每一个派生出来的表都必须有一个自己的别名）。

delete from stock   
where (org_id,material_id,state) 
in  (SELECT * from 
(select org_id,material_id, state from stock WHERE state = 1 group by org_id,material_id,state having count(*) > 1)
) 

当执行这条sql语句的时候就会出现Every derived table must have its own alias

(select org_id,material_id, state from stock WHERE state = 1 group by org_id,material_id,state having count(*) > 1)

就会产生一张新的表，和前面的表stock联合查询，但是mysql要求每一个派生出来的表都必须有一个自己的别名，那我给派生表加上别名即可；

eg：修改后的sql，直接在新生产的表中加入 他的别命名就行（“as a”或者“a”），“a”为新表的别名,这样就解决问题了

delete from stock   
where (org_id,material_id,state) in  
(SELECT * from (select org_id,material_id, state 
from stock WHERE state = 1 group by org_id,material_id,state having count(*) > 1) as a )