I found this code on www.databasejournal.com and it looks like it will do
what I want. However this code ages from 12/31/1997. What I need it to do,
is age from the currently selected member in the cube.

This code is currently constructing a shared dimension.
Any ideas?

IIF(DateDiff('d', "time_by_day"."the_date", '12/31/1997')< 30,
29,IIF(DateDiff('d', "time_by_day"."the_date", '12/31/1997')>= 30 AND
DateDiff('d',"time_by_day"."the_date", '12/31/1997')<= 60, 60,
IIF(DateDiff('d',"time_by_day"."the_date", '12/31/1997')>= 61 AND
DateDiff('d',"time_by_day"."the_date", '12/31/1997')<= 90, 90,
IIF(DateDiff('d',"time_by_day"."the_date", '12/31/1997')>= 91 AND
DateDiff('d',"time_by_day"."the_date", '12/31/1997')<= 120, 120, 999))))

Clint

Hi Clint -

The thing to do is to get the name of the time currentmember, and then in
another calc member convert it to a date so datediff can work. As long as
your member names for your day members are something that can be converted by
a cdate(), you should be OK.

Something like:

with member [measures].[CurrTimeName] as '[Time].CurrentMember.Name'

member [measures].[DaysDiff] as
'DateDiff("d",CDate([measures].[CurrTimeName]), Now())', solve_order = 10

member [measures].[Age] as 'IIF([measures].[DaysDiff] > 30,30, etc...)'
solve_order = 20

The above example will get the difference between the selected time member
and today.

You can combine these steps if you like, I just thought splitting them out
would make it easier to understand.

Good luck.

- Phil


"AshVsAOD" wrote:

Ah...

Once again, you solve my problem. Thanks!

That makes sense.
"SQL McOLAP" <SQLMcOLAP (AT) discussions (DOT) microsoft.com> wrote