MDX计算得出的度量，得出不同值的总和

Question

I have weird problem which I'm trying to resolve. My fact table looks like follow:

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

What I'm trying to do is create 2 measures:
Sum(ToPay)
Sum(Paid)

What is the problem?

In the result set I should get values like this: Client: 1 To Paid: 3000, Paid: 1250

Which means that sum(ToPay) will be calculated as distinct values for specific client and case.

It is possible to create such query in MDX ? If yes then how?

Answer 1

Ok - I tried to model your situation with the MS cube AdvWrks

Here is the situation where all are summed:

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];

Result is 500.

To pull out the sum of just distinct values I've used Filter , it depends on the set being in order so also needed to use the Order function:

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];

Result is 350.

Answer 2

If you do take Greg Galloway's advice, great for you. Otherwise, another good MDX challenge!

Here's what you can do.

First sort the set in its natural order.

Next, count the measure [To Pay] only for those cases where either the current member in the set is not the same as the previous member OR if it is the very first member(in which case the current will always not be the same as previous member)

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]

From Adventure Works:

    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]

Hope it works.