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Statement works in Access 2000 but not Access 2003

The following statement works in Access 2000, but when I run the same
database in Access 2003, I get a #Num! error.

=IIf(IsError([Reports]![rptDIAppActivityRevised]![rptHealthPaidRevised]![HPCount]),0,[Reports]![rptDIAppActivityRevised]![rptHealthPaidRevised]![HPCount])/IIf(IsError([Reports]![rptDIAppActivityRevised]![rptHealthApprovedRevised]![HACount]),1,[Reports]![rptDIAppActivityRevised]![rptHealthApprovedRevised]![HACount])

Any ideas?

Thanks,

Brian

0
bdaoust (130)
7/11/2005 6:24:21 PM
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BerkshireGuy <bdaoust@yahoo.com> wrote:
> The following statement works in Access 2000, but when I run the same
> database in Access 2003, I get a #Num! error.
>
> =IIf(IsError([Reports]![rptDIAppActivityRevised]![rptHealthPaidRevised]![HPCount]),0,[Reports]![rptDIAppActivityRevised]![rptHealthPaidRevised]![HPCount])/IIf(IsError([Reports]![rptDIAppActivityRevised]![rptHealthApprovedRevised]![HACount]),1,[Reports]![rptDIAppActivityRevised]![rptHealthApprovedRevised]![HACount])
>
> Any ideas?

I've had a few things not work when going from 2000->2003.....

#Num! suggests an arithmetic problem (as is #Div!).

Seperate out each part of your statement and display them in seperate 
controls. Check the values and make sure they are valid.
-- 
regards,

Bradley

A Christian Response
http://www.pastornet.net.au/response 


0
Br
7/11/2005 9:35:45 PM
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