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IFF expression

Posted on 2012-03-19
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Last Modified: 2012-06-27
Why is only the first part working on this?  This only subtracts 5% from all prices.

Iff([DatePosted]>3/4/2012,[Price]-.05*[Price],[Price]-.03*[Price])

I appreciate your help.
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Question by:ocdc
[X]
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6 Comments
 
LVL 75
ID: 37740764
Try this ... with #'s wrapping the date:

Iff([DatePosted] > #3/4 /2012#, [Price] - 0.05 * [Price], [Price] - 0.03 * [Price])

mx
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Author Comment

by:ocdc
ID: 37740790
the #'s prompt me for a date.  [DatePosted] is the date, so I should not be getting a prompt.
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LVL 75

Accepted Solution

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DatabaseMX (Joe Anderson - Microsoft MVP, Access and Data Platform) earned 2000 total points
ID: 37740793
I had one extra space in the date literal ... try this:

Iff([DatePosted] > #3/4/2012#, [Price] - 0.05 * [Price], [Price] - 0.03 * [Price])

or

Iff([DatePosted] > CDate("3/4/2012"), [Price] - 0.05 * [Price], [Price] - 0.03 * [Price])

Are you using this in a query?

mx
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Author Closing Comment

by:ocdc
ID: 37740798
It worked great.  Thanks.
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LVL 75
ID: 37740803
which one ?
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LVL 48

Expert Comment

by:Dale Fye
ID: 37741708
I'm assuming you guys meant  IIF( )   not    IFF( )

;-)

You could also have used:

[Price] * (1-iif([DatePosted] > #3/4/2012#, .05, .03))
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