Aggregating two dimension members

Hi to everybody!

First of all I'd like to apologize for my english!
...Then, this is the problem of a newbie of Sql Server Analysis Services, cubes and so on...,
I've created a cube, and I can easily browse it: what I wanted to obtain is a sort of Profit and Loss statement: dimensions are Revenues, Costs and periods, while "amount" is the only measure. It works!
But my question is: how can I aggregate two member of a same dimension? I.e.: how to create a new "row" in my cube where there are cost A plus cost B?
Does it make sense?

Thanks a lot

Jane Mischis

you can create a DIMENSION calculated member in the CALCULATIONS tab and have its parent hierarchy refer to COST hierarchy in the cost dimensions

in the expressions put

(<uniquename of cost A> + <uniquename of cost B>)

the new calcultion will appear as an additional row in the cost dimension.

hope this helps

|||Thank you so much for the answer..

Actually, what you suggest me is what I had done before writing here, but it didn't work, because:
- when I browse the cube, the new calculation does appear as an additional row in my members cost dimension, but it's transparent, like if it was hidden... and when I try to drag and drop it in row fields column on the right, I can't do it. I can just drop it above, in the filter box...

What's wrong?

Thank you,

Jane Mischis
|||

did you try doing it on a different browser (example Proclarity), cos the default browser that comes with analysis services can't perform all jobs.

|||No I didn't try, but I can't try, because I don't have it and I can't think I need it to solve my problem.
But I think the problem is elsewhere, maybe defining what dimensions and mesaures are.
The most important dimension is really "costs", but just the name of them, numbers come out from measure "amount"... And if I try to add a calculated Measures member, it works (i.e. "amount/1000")... but what I'd like to reach is just aggregate two names and having related sum i amounts.

Maybe I've to study a lot more...

Thank you however, Christina

Jane

|||Oh, great, I solve it.
I had just to put "all" in father member...

Thanks again

Jane