January 18, 2009 at 8:32 pm
I have a location dimension and in it I have region and site attribute. One region can have multi-sites
Now I want to union two sets like
Set 1:
region1 50
region2 30
Set 2:
site1 24
site2 26
site3 17
site4 13
And in the final result I want it looks like
region1 50 1
region2 30 1
site1 24 2
site2 26 2
site3 17 2
site4 13 2
Here the third column is to indicate it's region or site
Does anyone can help me for this?
January 19, 2009 at 5:14 am
Hi xyzsuper,
if I want to have the MDX-Code of any query, I usually display my cube through a end-user-client. I check "Options - Display MDX-Code" an then I copy and paste the code wherever I want to. You can download a trial-verion here: http://www.rbic.de/dwee
Regards
Ralf
January 19, 2009 at 5:16 am
Hi,
If you already have the two sets,
then you can 'union' them like this :
Union(Set1,Set2)
or
Set 1 + Set 2
or
{Set1,Set2}
Franky
Franky L.
January 19, 2009 at 10:36 am
Ok first, you need to remember that you can't create a "set" of members from more than one attribute hiearchy. Logically attributes are "crossed" to create results.
However, you can combine attributes to create a hierarchy. While this might seem to violate the above rule, think of it as creating a third thing, which you're going to use.
Create an attribute hierarchy, something like [Location].[by Region]. Put the region attribute as the first level, then the Site attribute as the second level.
Now, on to your sets.
The first set could either be defined as the list of the individual members { [Location].[by Region].&[region1], [Location].[by Region].&[region2] } or more compactly as [Location].[by Region].[Region].Members. The second set could also be compactly defined as [Location].[by Region].[Site].Members. Creating the union of these two sets is simply { [Location].[by Region].[Region].Members, [Location].[by Region].[Site].Members }. This is valid, because both sets are part of [by Region].
For your "third column", simply create a measure (i.e. [RegionLevel] ):
[Location].[by Region].CurrentMember.Level.Ordinal
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