【发布时间】:2016-04-01 00:36:31
【问题描述】:
如果我在选择中添加自定义列,我会收到此错误:
20018 列名“2”无效。
这里是查询示例
SELECT
[msg].[MessageTo],
[msg].[MessageFrom],
[msg].[SendTime],
[msg].[ReceiveTime],
[msg].[id],
'2' AS source,
[kat].[id] AS [CategoriId],
[kat].[naziv] AS [CategoriName]
FROM
[SMSServer_1].[dbo].[MessageIn] AS [msg]
LEFT JOIN [Tekijanka].[dbo].[crm_poruka] AS [por] ON [por].[fk_poruka] = [msg].[id]
AND [por].[fk_source] = [2]
LEFT JOIN [Tekijanka].[dbo].[crm_kategorije_poruka] AS [kat] ON [kat].[id] = [por].[fk_kategorija]
WHERE
msg.id NOT IN (
SELECT
fk_poruka
FROM
Tekijanka.dbo.crm_poruka
WHERE
fk_status <> 1
)
ORDER BY
[SendTime] DESC
有什么办法可以解决吗?
【问题讨论】:
-
AND [por].[fk_source] = [2]?任何表中都没有列名[2]。可能你想要AND [por].[fk_source] = 22 是一个整数值,[2] 是标识符 -
和 por,fk_source = 2 。 "por" 是表 "crm_poruka"....我在加入条件下需要它
-
删除方括号。没有名为
[2]的列。 -
我使用 laravel eloquent,它会自动添加方括号,我强制它不添加,现在它可以工作了......非常感谢,你帮助了我:)
标签: sql sql-server sql-server-2008 tsql laravel