MDX: Counting number of members of other dimension

[DominicB]

Hi,

I have following problem:

In my query I have 2 time dimensions and they are nested as follows

[Time1] [Time2]
01/04/2009
03/03/2009
04/03/2009
....

I need to count the number of members of the second time dimension
included in every day of the first dimension (this is needed to
calculate an average on a measure). I thought something like this
would work, but it always returns the total number of days in the
nested dimension:

([Time2].[Gregorian Calendar].[Date]
,[Time1].[Gregorian Calendar].currentmember).count

I'm sure this is a frequent requirement, but for the life of me, I
haven't been able to find anything working.

Thanks for your help!

[Zheng Xie]

From your result, [Time1] and [Time2] are two independent dimensions,
so the [Time1].currentmember would have no any effect on [Time2].

If don't change design, I think you maybe have to use vba's date
function to filter members of [Time2] based on name of
[Time1].currentmember.