Re: How to identify set (similar as possible)

currentmember always return a member, not a set.

what you don't know is if the dimension is currently used in a set!
using the extract function on a set when you don't know if the set contain
this dimension or not raise an error when the dimension is not here.
try to use the iserror function or convert the set to string and test the
string himself using string comaprison functions.



"igorkov" <igorkov@xxxxxxxxxxxxxxxxxxxxxxxxx> wrote in message
news:CBF79ADB-8CD8-4CDE-A2D1-59CFF7F88A11@xxxxxxxxxxxxxxxx
"Jéjé",

the problem is that you never know either is CurrentMember is a member or
set (lets assume). So i'd like to create a construction something like
this

IIF( [Time].[Fiscal].CurrentMember is set, IsLeaf(Extract.....),
IsLeaf([Time].[Fiscal].CurrentMember)

Thank you.

"Jéjé" wrote:

CurrentMember can be used against 1 dimension only, not against a set.

use the Extract function to extract 1 dimension only from your set, then
use
the currentmember
IsLEaf(Extract(<<my set>>, [Time].[Fiscal]).CurrentMember)





.

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