Thread: [Solved] Using DateDif Function
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Old 03-14-2014, 01:26 AM
OTPM OTPM is offline Windows 7 32bit Office 2010 32bit
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Hi Julie
Many thanks for your prompt response (yes I did spell the Function name wrong :-)).
I ended up using this formula in the end but will also try your solution:

IIf([Finish]>Date(),IIf([Finish]-Date()<=3,"TRUE","FALSE"),"NA")

Edit:
I modified your solution in an attempt to ignore those tasks that were complete but the edit does not work, any ideas?

IIf([Status]="Complete","",IIf(DateDiff("d",[Finish],[Current Date])>0 And DateDiff("d",[Finish],[Current Date])<=3,True,False))

Thanks

Tony
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