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Dave Peterson
 
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Default how can I fix a #value! error?

Well, you wouldn't want a plus there. But maybe:

=IF(INDIRECT("'xyz'!j3")=3,O13*I6,IF(INDIRECT("'xy z'!j3")=8,O14*I6,1))


Dave Peterson wrote:

if xyz!j3=3
then 013*i6
else 1

But then you have another completely separate If statement there.

Did you want
=IF(INDIRECT("'xyz'!j3")=3,O13*I6,1) + IF(INDIRECT("'xyz'!j3")=8,O14*I6,1)

reloanpro wrote:

I get a #VALUE! with the following formula:

"=IF(INDIRECT("'xyz'!j3")=3,O13*I6,1),IF(INDIRECT( "'xyz'!j3")=8,O14*I6,1),..."

The logic seems to work, as values are correct when I evaluate the
calculation steps.

Any help is appeciated.


--

Dave Peterson


--

Dave Peterson