Hello,
I have a Time hierarchy (FY - Period Num) as follows:
- Fiscal Year
- - Period Number
I am trying to get the number of just PeriodNumber 1 for each year for Actives. I have tried the following:
([Measures].[Active] , OpeningPeriod( [Time].[FY - Period Num].[Time].[FY - Period Num].currentmember ) )
That returns nothing.
([Measures].[Active] , OpeningPeriod( [Time].[FY - Period Num].[Period Number] , [Time].[FY - Period Num].currentmember ) )
That just returns the current period number.
I had also tried to do stuff like:
([Measures].[Active] , [Time].[Period Number] .currentmember)
But that again only gave me the current period data
([Measures].[Active] , [Time].[Period Number] .[&1])
Gives me a #VALUE! error
I also tried:
([Measures].[Active] , [Time].[Period Number] .[&1], [Time].[Fiscal Year] .currentmember)
I can do this:
([Measures].[Active] , [Time].[FY - Period Num].[Period Number].[&1]&[2006])
Here I get data, but it is only for 2006.
And on, and on. Obviously, I don't know what I am doing, but I think I am close.
How can I get the first period data from the current year?
Thank you.
-Gumbatman
Hello Again GumbatMan.
The answere to this question depends on if you have a time dimension that streches further away than the current year.
If you only have the current year as the last member on the year level you could write MDX like:
YourTimeDimension.(FY - Period Num).LastChild.FirstChild.
This will give you the first child member of the last year in your time dimension/hierarchy.
In the enterprise edition of SSAS2005 you have aggregation functions that will take care of this problem. Have a look at the properties for the measure in the relevant measure group.
Here is a link to another approach provided that you are looking for the current time member in reality:
http://blogs.conchango.com/christianwade/archive/2006/06/23/MDX-Script_3A00_-Current_2F00_Relative-Period.aspx
Think about if your problem is about aggregations(key wordemi additive measure) or the time member on the column or row axis in a client.
HTH
Thomas Ivarsson
|||Thomas,
Once again you've been very, very helpful.
Thank you!
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