Question

我使用以下内容：

WITH 
  SET [myset] AS 
    Order
    (
      {
          [Customer].[Country].[Country].MEMBERS
        * 
          [Customer].[Customer].[Customer].MEMBERS
      }
     ,[Measures].[Internet Sales Amount]
     ,ASC
    ) 
  MEMBER [Measures].[rank] AS 
    Rank
    (
      (
        [Customer].[Country].CurrentMember
       ,[Customer].[Customer].CurrentMember
      )
     ,[myset]
    ) 
  MEMBER [Measures].[newrank] AS 
    IIF
    (
        [myset].Item(
        [Measures].[rank] - 1).Item(0).Name
      <> 
        [myset].Item(
        [Measures].[rank] - 2).Item(0).Name
     ,1
     ,
      ([myset].Item([Measures].[rank] - 2),[Measures].[newrank]) + 1
    ) 
SELECT 
  {
    [Measures].[Internet Sales Amount]
   ,[Measures].[rank]
   ,[Measures].[newrank]
  } ON 0
 ,[myset] ON 1
FROM [Adventure Works];

这是Amish Shah的作品：http://blog.sqltechie.com/2010/03/rank-with-partitioning-mdx.html

目前它没有给出具有相同排名的元组的元组，例如如果x的互联网销售额为10且y也有10，则在上面它们的排名不会相同。

上述内容是否可以更改，以便它们具有相同的值，那么它们对[Measures].[rank]具有相同的值 - 并且[Measures].[newrank]的值也相同？

（在[Measures].[rank]的定义中添加第三个参数似乎不是一个可行的路径，因为Rank函数然后进入逐个单元计算模式）

Answer 1

如何使用递归计算成员 -

WITH 
  SET [myset] AS 
    Order
    (
      {
          [Customer].[Country].[Country].MEMBERS
        * 
          [Customer].[Customer].[Customer].MEMBERS
      }
     ,[Measures].[Internet Sales Amount]
     ,ASC
    ) 
  MEMBER [Measures].[myrank] AS 
    Rank
    (
      (
        [Customer].[Country].CurrentMember
       ,[Customer].[Customer].CurrentMember
      )
     ,[myset]
    ) 


member prevmembervalue as 
(myset.item(myRank - 2), [Measures].[Internet Sales Amount])

member currentmembervalue as
(myset.item(myRank - 1), [Measures].[Internet Sales Amount])

MEMBER greaterthanprev as
iif(currentmembervalue > prevmembervalue, 1, 0)

member rankActual as
iif
(
 prevmembervalue = null, 1,
 iif(currentmembervalue > prevmembervalue,
  (myset.item(myRank - 2), rankActual) + 1,
  (myset.item(myRank - 2), rankActual)
 )
 )

select myset on 1,
{[Measures].[Internet Sales Amount], rankActual, myRank}
on 0
from 
[Adventure Works]

实际上不需要greaterthanprev，currentmembervalue和prevmembervalue这些措施。在这里添加它们只是为了进一步澄清过程。

绑定排名的帐户

1 个答案: