I have 2 cubes with 4 and 3 dimensions each.
I have created a virtual cube and added all the 4 dimesnions to the virtual
cube.

When the 4th dim is at All, the browser shows all measures, when a value is
selected on the 4th dim, the measures from the 2nd cube disappear.

Can I not have "unshared" dims in a virtual cube? Is it only a problem with
viewing the measure in the browser?

thanks

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Message posted via http://www.sqlmonster.com

The measure that goes missing is a calculated measure in the source cube.
I have tried to create another calc measure in the virtual cube as
validmeasure([measure].[measurename]), but that also does not seem to work

Please help!!

thanks

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Message posted via http://www.sqlmonster.com

Angie;

In order to use Virtual Cubes, you need to have "shared" dimensions that
exist in both cubes you want to join. The reason why your measures can
disappear on you is because there are no measure records for the dimensions
that you chose. For instance, if you have a product dimension and an employee
dimension, but they both do not exist in both cubes, then you will be unable
to see product hours or employee sales. The virtual cube will not make up
records for the dimensions that exist in one cube, there has to be
commonality amongst the data to make the join. Depending on the cubes you
want to join, depends on how many dimensions you can use. The more "shared"
dimensions you have in the two cubes, then the better the analysis can be.

A dimension that is not shared, cannot be used in a virtual cube, as there
is no chance that the records associated to the dimension will exist in the
cube you are joining it to.

Let me know how it goes.

"angie via SQLMonster.com" wrote:

Jonathan,

I just used a calc measure as below. validmeasure function can also be used
- but i seem to be making a mistake while using it - it does not work for
me - but the below worked well!

([Measures].[measurename],[Missing dim], [All])

thank you

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Message posted via http://www.sqlmonster.com

What is the error, or unexpected behaviour you are getting?

When you use the formula, do you see ANY results?


"Brian via SQLMonster.com" wrote: