DateDiff function 1

[alectek]

Hi!

{StartDate} = 07/01/2003
and {EndDate} = 01/31/2004

then the DateDiff("m", {StartDate}, {EndDate}) = 6

But, really it is 7.

I can use formula like
Round (({EndDate} - {StartDate})/30) and result will be 7.
Maybe somebody can give me more elegant solution.

Thank you very much.
Alex.

 

[Naith]

Try DateDiff("m", {StartDate}, {EndDate}) + 1

Naith

 

[Reebo99]

How many months would you expect :

{StartDate} = 07/31/2003
and {EndDate} = 01/01/2004 ??







Reebo
UK

Please Note - Due to current economic forcast and budget constraints, the light at the end of the tunnel has been switched off. Thank you for your cooperation.

 

[alectek]

To Reebo.
{EndDate} = 01/31/2004 and I expect 7 month? And you?

Thanks.
Alex.

 

[Reebo99]

sorry, I mean If the dates were from 31st July 2003 to 1st January 2004, how many months would you expect?

Personally, when showing the difference in months I don't take into account the day. Therefore :

1st July 2003 to 31st January 2004 = 6 Months
31st July 2003 to 1st January 2004 = 6 Months

If you want 1st July 2003 to 31st January 2004 = 7 Months then 31st July 2003 to 1st January 2004 = 5 Months ?






Reebo
UK

Please Note - Due to current economic forcast and budget constraints, the light at the end of the tunnel has been switched off. Thank you for your cooperation.

 

[jymm]

wowie --- must resist red flag of last post --- must resist

I ran into something similar a while back and it seems that the datediff was truely counting the number between - not including the end ( so - July, August, September, October, November, December was 6).

Knowing that I used it the way I wanted --- can't find the reference for that right now (sorry).

 

[alectek]

To Naith!
Thank you very much. Very easy solution and it's working even for one month.

Alex.