DateTime usage for duration computation

Marsel has asked for the wisdom of the Perl Monks concerning the following question:

Dear Monks,

I need to compute duration intervals from various dates. I read through the DateTime Tutorial on perlmonks and choose to use this module. Unfortunately, i'm getting strange results on a first test : misunderstanding or bug ?

here is the code :

use strict;
use warnings;
use DateTime;

my $t1 = DateTime->new(year => 2001,
                        month => 4,
                        day => 23,
                        hour => 22,
                        minute => 32);
my $t2 = DateTime->new(year => 2001,
                        month => 4,
                        day => 22,
                        hour => 15,
                        minute => 42);
print "T1 = ".$t1->hour.":".$t1->minute."\n
       T2 = ".$t2->hour.":".$t2->minute."\n
       T1-T2 = ".$t1->subtract_datetime($t2)->hours."\n
       T2-T1 = ".$t2->subtract_datetime($t1)->hours."\n";
[download]

Here is the result :

T1 = 22:32
T2 = 15:42
T1-T2 = 6
T2-T1 = 6

Any comment or help would be appreciated, or if you know a better module to do that (i'll need to compute duration intervals in term of days also, which implies knowing the month's lengh ...)

Thanks,

Marcel

Comment on DateTime usage for duration computation Download Code

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Re: DateTime usage for duration computation by derby (Abbot) on Aug 29, 2007 at 12:25 UTC
I think it's part misunderstanding and part bad documentation. Looking at DateTime, the subtract_datetime method returns a DateTime::Duration object. Looking at DateTime::Duration hours method, it's a a wrapper over the in_units method which when asking for hours, does an int on the numbers, effectively rounding the value (but probably not in a way you want). -derby update: I prefer Date::Manip for date calculations.	[reply]
Re: DateTime usage for duration computation by andreas1234567 (Vicar) on Aug 29, 2007 at 13:04 UTC
DateTime::Duration: * years, months, weeks, days, hours, minutes, seconds, nanoseconds These methods return numbers indicating how many of the given unit the object represents, after having done a conversion to any larger units. For example, days are first converted to weeks, and then the remainder is returned. These numbers are always positive. $ perl -l use strict; use warnings; use DateTime; my $t1 = DateTime->new( year => 2001, month => 4, day => 23, hour => 22, minute => 32 ); my $t2 = DateTime->new( year => 2001, month => 4, day => 22, hour => 15, minute => 42 ); # subtract_datetime returns a "DateTime::Duration" object my $dtd1 = $t1->subtract_datetime($t2); print $dtd1->years . q{ years} if $dtd1->years; print $dtd1->months . q{ months} if $dtd1->months; print $dtd1->weeks . q{ weeks} if $dtd1->weeks; print $dtd1->days . q{ days} if $dtd1->days; print $dtd1->hours . q{ hours} if $dtd1->hours; print $dtd1->minutes . q{ minutes} if $dtd1->minutes; print $dtd1->seconds . q{ seconds} if $dtd1->seconds; print $dtd1->nanoseconds . q{ nanoseconds} if $dtd1->nanoseconds; __END__ 1 days 6 hours 50 minutes [download] -- Andreas	[reply] [d/l]
Re: DateTime usage for duration computation by girarde (Hermit) on Aug 29, 2007 at 14:09 UTC
I prefer `Date::Manip` as well. This is a case, though, where could go with something like: `my $t1 = DateTime->new(year => 2001, month => 4, day => 23, hour => 22, minute => 32); my $t2 = DateTime->new(year => 2001, month => 4, day => 22, hour => 15, minute => 42); my $delta = $t1->subtract_datetime_absolute($t2) my $days = int($delta / 86400); $delta = $delta % 86400; my $hours = int($delta / 3600); $delta = $delta % 3600);` [download] etcetera. Not elegant, but nice and obvious what's happening.	[reply] [d/l] [select]
Re^2: DateTime usage for duration computation by varanasi (Scribe) on Oct 20, 2014 at 01:28 UTC
I think that your $delta, since it's a duration object, needs to be dereferenced into seconds to make this work. `my $days = int($delta->seconds / 86400);` Right?	[reply] [d/l]