DateDiff Function

B

Bob Watts

Where [Finish] = 9/10/03 and [Finish10] = 9/9/03;
I expect the following to yield the value 1:
DateDiff("d",[Finish],[Finish10],).

Where is there better definition of this function and what
the heck am i doing wrong? Are there examples anywhere?
 
D

Dale Howard

Bob --

To solve your problem, you might try using the following formula in a custom
Duration field:

ProDateDiff([Finish],[Finish10])

I believe it will yield the answer that you seek. If you don't like seeing
a negative Duration value, you can also apply the Absolute Value function to
the formula as follows:

Abs(ProjDateDiff([Finish],[Finish1]))

Hope this helps.
 
B

Bob Watts

Dale--
Thanks.

I used ProjDateDiff([Finish],[Baseline1 Finish]) where:
Finish=9/25/03 and Baseline1 Finish = 9/26/03

Expected result = 1 Day
Actual result when custom field described as
type: "Number" = 480
Actual result when custom field described as
type: "Duration" = 1 Day!

Thanks, again
-----Original Message-----
Bob --

To solve your problem, you might try using the following formula in a custom
Duration field:

ProDateDiff([Finish],[Finish10])

I believe it will yield the answer that you seek. If you don't like seeing
a negative Duration value, you can also apply the Absolute Value function to
the formula as follows:

Abs(ProjDateDiff([Finish],[Finish1]))

Hope this helps.




Bob Watts said:
Where [Finish] = 9/10/03 and [Finish10] = 9/9/03;
I expect the following to yield the value 1:
DateDiff("d",[Finish],[Finish10],).

Where is there better definition of this function and what
the heck am i doing wrong? Are there examples anywhere?
Click to expand...


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