what happens if you do
#difDate#
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There isn't an quick and easy solution, although a few straight forward ones.
Everyone is going to suggest you do a DateDiff on each date part you want to
handle, but after quickly glancing at cflib.org, think that a better solution
might be the following.
Use DateDiff to get the difference in
Well you're going to need to do it in parts, right, so given the dates
in your example,
1/1/2007 - 2/9/2008 returns 1 year, 1 month & 8 days
So ...
startdate = "1/1/2007";
enddate = "2/9/2008";
days = datediff("d",startdate,enddate);
years = int(days / 365);
months = int(days / 30) - (12
I am trying to figuring how to determine the year, months and days from a date.
Like from 1/1/2007 to 2/9/2008 it is 1 year 1 month and 8 days. I am trying to
figure out how to do that programmatically.
I have been trying to do it with datediff but I cannot figure out how.
Any help would be gre
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