RANK()Over Partition BY不工作

Question

当我运行下面的代码时,ROWID始终为1.我需要ID为1,对于具有相同信用值的每个项目.

;WITH CTETotal AS (SELECT
     TranRegion
    ,TranCustomer
    ,TranDocNo
    ,SUM(TranSale) 'CreditValue'
FROM dbo.Transactions

LEFT JOIN customers AS C 
      ON custregion = tranregion 
      AND custnumber = trancustomer
LEFT JOIN products AS P
      ON prodcode = tranprodcode

GROUP BY
TranRegion
,TranCustomer
,TranDocNo)

SELECT
       r.RegionDesc
      ,suppcodedesc
      ,t.tranreason as [Reason]
      ,t.trandocno as [Document Number]
      ,sum(tranqty) as Qty
      ,sum(tranmass) as Mass
      ,sum(transale) as Sale
      ,cte.CreditValue AS 'Credit Value'
      ,RANK() OVER (PARTITION BY cte.CreditValue ORDER BY cte.CreditValue)AS ROWID

FROM transactions t

LEFT JOIN dbo.Regions AS r    
      ON r.RegionCode = TranRegion  

LEFT JOIN CTETotal AS cte
      ON cte.TranRegion = t.TranRegion
      AND cte.TranCustomer = t.TranCustomer
      AND cte.TranDocNo = t.TranDocNo

GROUP BY 
       r.RegionDesc
      ,suppcodedesc
      ,t.tranreason
      ,t.trandocno
      ,cte.CreditValue

ORDER BY CreditValue ASC

编辑

所有400的信用值必须将ROWID设置为1.并且所有200的信用值必须将ROWID设置为2.依此类推.

Answer 1

你需要这样的东西吗？

with cte (item,CreditValue)
as
(
select 'a',8 as CreditValue union all
select 'b',18 union all
select 'a',8 union all
select 'b',18 union all
select 'a',8 
) 
select CreditValue,dense_rank() OVER (ORDER BY item)AS ROWID from cte

结果

CreditValue ROWID
----------- --------------------
8           1
8           1
8           1
18          2
18          2

在你的代码替换

,RANK() OVER (PARTITION BY cte.CreditValue ORDER BY cte.CreditValue)AS ROWID 

通过

,DENSE_RANK() OVER (ORDER BY cte.CreditValue)AS ROWID