Question

我尝试用这个选择计算来自另一个表的所有项目：

SELECT id, name,  (SELECT count(*) 
                   FROM prekes_main 
                   WHERE prekes_main.pristKaina = 1 
                   and   prekes_main.pg_kodas LIKE 'grupes_main.pg_kodas%') as pristKaina
FROM grupes_main 
WHERE grupes_main.level = 1 
and grupes_main.name <> ''

在LIKE子句中我想自动选择grupes_main列pg_kodas，但在这个查询中它总是返回0，LIKE函数中的错误在哪里？ THX

Answer 1

SELECT  id, name,
        (
        SELECT  COUNT(*)
        FROM    prekes_main
        WHERE   prekes_main.pristKaina = 1
                AND prekes_main.pg_kodas LIKE CONCAT(grupes_main.pg_kodas, '%')
        ) pristKaina
FROM    grupes_main
WHERE   grupes_main.level = 1
        AND grupes_main.name <> ''

在sql LIKE子句中哪里出错？

1 个答案: