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Nz() in IIF |
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| message from =?Utf-8?B?Sk1vcnJlbGw=?= on 17 May 2004 |
=IIf([aupdateval],(nz([annbal]-[suma])+[aupdateval]),(nz([annbal]-[suma])))
If [aupdateval] contains a value and if there is no value for [suma], the statement only returnes [aupdateval]. What am I missing in my IIF statement? Ultimately, I want to return the sum of ([annbal]-[suma]+[aupdateval]), regardless of their initial values (null or not).
tia,
JMorrell
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| fredg replied to =?Utf-8?B?Sk1vcnJlbGw=?= on 17 May 2004 |
Does this work for you?
=IIf([aupdateval],Nz([annbal])-Nz([suma])+[aupdateval],
(Nz([annbal])-Nz([suma]))
Or more simply (no IIf needed):
=Nz([annbal]) - Nz([suma]) + Nz([aupdateval])
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| =?Utf-8?B?Sk1vcnJlbGw=?= replied to fredg on 17 May 2004 |
The simple solution is always the best. It works fine, thank you.
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Archived message: Nz() in IIF (Microsoft Access Database)
