Re: Code to calculate Delta b/w two dates in seconds

my $t1 = "2009-04-12 23:59:55 PM";
my $t2 = "2009-03-30 00:00:02 PM";
[download]

There is no 23:59:55 with PM. Either you got 24hour-format or AM/PM :-)

Your solution assums that each day has 24 hours. It's a common mistake, but it is simply not true. There is daylight saving, once they even put some days difference in a second and February hasn't always 28 days.

use Time::Local;

sub String2Time {
 $_[0] =~ /^(\d{4})\-(\d{2})\-(\d{2}) (\d{2})\:(\d{2})\:(\d{2})/ or di
+e "Invalid format: $_[0]";
 return timelocal($6,$5,$4,$3,($2 - 1),($1 - 1900));
}

my $t1 = "2009-04-12 23:59:55 PM";
my $t2 = "2009-03-30 00:00:02 PM";

print &String2Time($t1) - &String2Time($t2);
[download]

Just parse your timestamp and feed it into timelocal from Time::Local.

Please let us know if this is faster or slower than your solution.

Try Padre - the free Perl IDE

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Re^2: Code to calculate Delta b/w two dates in seconds by snra_perl (Acolyte) on Sep 22, 2009 at 22:26 UTC
Thanks for the inputs. It works well and speed of execution seems to have no much difference. Can we use the same approach of Time::Local to add a some seconds to the existing time. I need something to work like the behaviour mentioned below `my $t1 = "2009-04-12 23:59:55"; my $delta = 600; # seconds $updated_time = &String2Time($t1) + $delta;` [download] Thanks again	[reply] [d/l]