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Sql 是另一个-多部分标识符“;xxx“;无法约束_Sql_Sql Server 2008_Tsql - Fatal编程技术网

Sql 是另一个-多部分标识符“;xxx“;无法约束

Sql 是另一个-多部分标识符“;xxx“;无法约束,sql,sql-server-2008,tsql,Sql,Sql Server 2008,Tsql,好的,这是我的问题: SELECT DISTINCT CS.CPL_Schedule_Id AS CplScheduleId ,S.Schedule_Status_Id AS ScheduleStatusId ,S.Record_Created AS Created FROM CPL_Schedule CS JOIN ( SELECT TOP 1 CSL.CPL_Schedule_ID, C

好的,这是我的问题:

SELECT DISTINCT
    CS.CPL_Schedule_Id          AS CplScheduleId
    ,S.Schedule_Status_Id       AS ScheduleStatusId
    ,S.Record_Created           AS Created
FROM CPL_Schedule CS
    JOIN (
            SELECT TOP 1 CSL.CPL_Schedule_ID, CSL.Record_Created, CSL.Schedule_Status_Id
            FROM CPL_Schedule_Audit_Log CSL
            WHERE CSL.CPL_Schedule_ID = CS.CPL_Schedule_ID
            ORDER BY Record_Created DESC
        ) S
        ON S.CPL_Schedule_ID = CS.CPL_Schedule_ID
WHERE 
    CS.Exhibitor_Id = 1 
我在这行
上得到了错误,其中CSL.CPL\u Schedule\u ID=CS.CPL\u Schedule\u ID

无法绑定多部分标识符“CS.CPL\u Schedule\u ID”


对我来说,这似乎是说嵌套查询不知道CS,是真的吗?我错过了什么?非常感谢您的帮助,谢谢

看起来,您需要
交叉应用
而不是
加入
,因此不需要
上的

SELECT DISTINCT
    CS.CPL_Schedule_Id          AS CplScheduleId
    ,S.Schedule_Status_Id       AS ScheduleStatusId
    ,S.Record_Created           AS Created
FROM CPL_Schedule CS
    CROSS APPLY (
            SELECT TOP 1 CSL.CPL_Schedule_ID, CSL.Record_Created, CSL.Schedule_Status_Id
            FROM CPL_Schedule_Audit_Log CSL
            WHERE CSL.CPL_Schedule_ID = CS.CPL_Schedule_ID
            ORDER BY Record_Created DESC
        ) S
WHERE 
    CS.Exhibitor_Id = 1 

看起来,您需要交叉应用
而不是
加入
,因此不需要在
上使用

SELECT DISTINCT
    CS.CPL_Schedule_Id          AS CplScheduleId
    ,S.Schedule_Status_Id       AS ScheduleStatusId
    ,S.Record_Created           AS Created
FROM CPL_Schedule CS
    CROSS APPLY (
            SELECT TOP 1 CSL.CPL_Schedule_ID, CSL.Record_Created, CSL.Schedule_Status_Id
            FROM CPL_Schedule_Audit_Log CSL
            WHERE CSL.CPL_Schedule_ID = CS.CPL_Schedule_ID
            ORDER BY Record_Created DESC
        ) S
WHERE 
    CS.Exhibitor_Id = 1 

这太棒了,我怎么不知道交叉申请?!这太棒了,我怎么不知道交叉申请?!