Count(IIF) not working

Hello, I have been fighting with getting the following statement to work
=Count(IIF( Fields!PDDAYS.Value>30,1,0))
I basically have returned all records to get a total count and need to weed
out those I want so I can divide the [totalofallrecords] by the
[totalofthose>30daysdue]
Any help would be appreciated.
SonjaI suppose you really want the Sum() aggregate, rather than Count(). Count
works the same way as it works in SQL - every value different than NULL will
be counted.
Try this (inside a data region - e.g. in a table header/footer etc.):
=Sum(IIF( Fields!PDDAYS.Value>30,1,0))
-- Robert
This posting is provided "AS IS" with no warranties, and confers no rights.
"SLB" <SLB@.discussions.microsoft.com> wrote in message
news:1D0DCEDA-882C-41E5-9FB7-3308E91DF828@.microsoft.com...
> Hello, I have been fighting with getting the following statement to work
> =Count(IIF( Fields!PDDAYS.Value>30,1,0))
> I basically have returned all records to get a total count and need to
> weed
> out those I want so I can divide the [totalofallrecords] by the
> [totalofthose>30daysdue]
> Any help would be appreciated.
> Sonja|||Thanks so much, worked like a charm. Don't know what I was thinking.
"Robert Bruckner [MSFT]" wrote:
> I suppose you really want the Sum() aggregate, rather than Count(). Count
> works the same way as it works in SQL - every value different than NULL will
> be counted.
> Try this (inside a data region - e.g. in a table header/footer etc.):
> =Sum(IIF( Fields!PDDAYS.Value>30,1,0))
>
> -- Robert
> This posting is provided "AS IS" with no warranties, and confers no rights.
>
> "SLB" <SLB@.discussions.microsoft.com> wrote in message
> news:1D0DCEDA-882C-41E5-9FB7-3308E91DF828@.microsoft.com...
> > Hello, I have been fighting with getting the following statement to work
> >
> > =Count(IIF( Fields!PDDAYS.Value>30,1,0))
> >
> > I basically have returned all records to get a total count and need to
> > weed
> > out those I want so I can divide the [totalofallrecords] by the
> > [totalofthose>30daysdue]
> >
> > Any help would be appreciated.
> > Sonja
>
>