Question

我需要创建一个物料主文件列表，其中将显示最新的发票和销售订单记录。我创建的代码显示了所有记录，我需要一个带有最新DocDate的记录。...

非常感谢您的协助！

        SELECT DISTINCT
        T0.[ItemCode], 
        T1.[ObjType], 
        T1.[DocNum], 
        MAX(T1.[DocDate])  AS 'LastActivity'
    FROM INV1 T0  INNER JOIN OINV T1 ON T0.[DocEntry] = T1.[DocEntry]


    GROUP BY 
        T0.[ItemCode], 
        T1.[ObjType], 
        T1.[DocNum]

UNION ALL

    SELECT DISTINCT
        T0.[ItemCode], 
        T1.[ObjType], 
        T1.[DocNum], 
        MAX(T1.[DocDate])  AS 'LastActivity'
    FROM RDR1 T0  INNER JOIN ORDR T1 ON T0.[DocEntry] = T1.[DocEntry]


    GROUP BY 
        T0.[ItemCode], 
        T1.[ObjType], 
        T1.[DocNum]

Answer 1

如果我理解正确，您可以使用UNION ALL将发票和订单表中的数据汇总在一起，然后然后进行汇总：

SELECT i.[ItemCode], o.[ObjType], o.[DocNum], 
       MAX(o.[DocDate])  AS LastActivity
FROM INV1 i JOIN
     ((SELECT DocEntry, ObjType, DocDate
       FROM OINV o
      ) UNION ALL
      (SELECT DocEntry, ObjType, DocDate
       FROM ORDR o
      )
     ) o
     ON i.[DocEntry] = o.[DocEntry]
GROUP BY i.[ItemCode], o.[ObjType], o.[DocNum];

Answer 2

添加到您的查询

ORDER BY LastActivity DESC  
OFFSET 0 ROWS  
FETCH NEXT 1 ROWS ONLY

所以您只会得到具有最新DocDate的行。
对于SQL Server 2012+。
编辑：
尝试将查询嵌入此代码中：

SELECT t.[ItemCode], t.[ObjType], t.[DocNum], MAX(t.LastActivity)
FROM (
  <your query>
) t
GROUP BY t.[ItemCode], t.[ObjType], t.[DocNum]

Answer 3

专家，在每个发票和销售订单中每个项目仅提供最新记录的情况下，这似乎是最好的方法。

SELECT A.ItemCode, A.ObjType, A.DocNum, A.LastActivity

从（选择T0。[ItemCode]，T1。[ObjType]，T1。[DocNum]，T1。[DocDate] AS'LastActivity'， ROW_NUMBER（）超过（按T0.ItemCode排序按T0.DocEntry DESC排序）作为'rownum' 从INV1 T0 T0。[DocEntry] = T1。[DocEntry]上的INNER JOIN OINV T1。 T0。[ItemCode] ='A00001_测试非库存'或T0。[ItemCode] ='A00001'

 UNION ALL

 SELECT T0.[ItemCode], T1.[ObjType], T1.[DocNum], T1.[DocDate] AS 'LastActivity',
     ROW_NUMBER() OVER (PARTITION BY T0.ItemCode ORDER BY T0.DocEntry DESC) AS 'rownum'
 FROM RDR1 T0
    INNER JOIN ORDR T1 ON T0.[DocEntry]=T1.[DocEntry]

    WHERE T0.[ItemCode] = 'A00001_ test non inventory' OR T0.[ItemCode] = 'A00001'
    ) A

A.rownum = 1;

获取每个项目的最新记录

3 个答案: