MDX计算得出的度量，得出不同值的总和

Question

我有一个奇怪的问题要解决。 我的事实表如下所示：

CaseID | ClientID | PaymentDate|ToPay   | Paid
1      |     1    | 2015-01-01 |1000    |100
1      |     1    | 2015-02-01 |1000    |200
1      |     1    | 2015-03-01 |1000    |300
2      |     1    | 2015-01-01 |2000    |100
2      |     1    | 2015-02-01 |2000    |400
2      |     1    | 2015-03-01 |2000    |150

我想做的是创建2个措施：
Sum(ToPay)
Sum(Paid)

问题是什么？

在结果集中，我应该获得如下所示的值：客户：1支付：3000，支付：1250

这意味着sum（ToPay）将被计算为特定客户和案例的不同值。

是否可以在MDX创建这样的查询？ 如果是，那怎么办？

Answer 1

好的-我尝试使用MS多维数据集AdvWrks对您的情况进行AdvWrks

这是所有情况相加的情况：

WITH 
  MEMBER [Date].[Calendar].[All].[x] AS 100 
  MEMBER [Date].[Calendar].[All].[y] AS 200 
  MEMBER [Date].[Calendar].[All].[z] AS 200 
  SET [blah] AS 
    {
      [x]
     ,[y]
     ,[z]
    } 
  MEMBER [Measures].[xx] AS 
    Sum
    (
      [blah]
     ,[Measures].[Internet Sales Amount]
    ) 
SELECT 
  {[Measures].[xx]} ON 0
FROM [Adventure Works];

结果是500。

为了提取我使用过Filter不同值的总和，这取决于设置的顺序，因此还需要使用Order函数：

WITH 
  MEMBER [Date].[Calendar].[All].[a] AS 100 
  MEMBER [Date].[Calendar].[All].[x] AS 50 
  MEMBER [Date].[Calendar].[All].[y] AS 200 
  MEMBER [Date].[Calendar].[All].[z] AS 200 
  SET [blah] AS 
    {
      [a]
     ,[x]
     ,[y]
     ,[z]
    } 
  MEMBER [Measures].[xx] AS 
    Sum
    (
      Filter
      (
        Order
        (
          [blah]
         ,[Measures].[Internet Sales Amount]
        ) AS x
       ,
          (
            x.CurrentMember
           ,[Measures].[Internet Sales Amount]
          )
        <> 
          (
            x.Item(
            x.CurrentOrdinal)
           ,[Measures].[Internet Sales Amount]
          )
      )
     ,[Measures].[Internet Sales Amount]
    ) 
SELECT 
  {[Measures].[xx]} ON 0
FROM [Adventure Works];

结果是350。

Answer 2

如果您接受Greg Galloway的建议，那对您非常有用。 否则，另一个很好的MDX挑战！

这是您可以做的。

首先按照自然顺序对集合进行排序。

接下来，算上措施[To Pay]只对那些情况下，无论是在设定的当前成员是不一样的前一个成员，或者如果它是第一个成员（在这种情况下，电流将永远不会是相同的前成员）

with set [Client&Case] as
order
     (
        [Fact].[ClientID].[All].children * [Fact].[CaseID].[All].children,
        [Fact].[ClientID].currentmember.name + "" + [Fact].[CaseID].currentmember.name
     )


member measures.[Actual - To Pay]  as 
sum
    (
     [Client&Case],
      iif
      (    
        not([Client&Case].item([Client&Case].currentordinal-1) is [Client&Case].item([Client&Case].currentordinal-2)) 
           --current not same not as previous
        or [Client&Case].currentordinal = 1 --the first member!
       ,[Measures].[To Pay]
       ,0
      )
    )

select 
{measures.[Paid], measures.[Actual - To Pay]} on 0,
[Client&Case] on 1
from [Your Cube]

从冒险作品：

    with set [Category&Country] as
    {
    ([Product].[Category].&[3], [Customer].[Customer Geography].[Country].&[Australia]),
    ([Product].[Category].&[3], [Customer].[Customer Geography].[Country].&[Australia]),
    ([Product].[Category].&[4], [Customer].[Customer Geography].[Country].&[Australia]),
    ([Product].[Category].&[4], [Customer].[Customer Geography].[Country].&[Australia]),
    ([Product].[Category].&[4], [Customer].[Customer Geography].[Country].&[Australia])
    }

    member measures.TotalInternetSales as
    sum([Category&Country],[Measures].[Internet Sales Amount])

    member measures.[ActualInternetSales]  as 
    sum
        (
         [Category&Country],
          iif
          (         
            not([Category&Country].item([Category&Country].currentordinal-1) is [Category&Country].item([Category&Country].currentordinal-2)) 
            or [Category&Country].currentordinal = 1
           ,([Category&Country].currentmember, [Measures].[Internet Sales Amount])
           ,0
          )
        )

    select 
    {measures.TotalInternetSales, measures.[ActualInternetSales]} on 0
    from [Adventure Works]

希望它能工作。