Subtract Hours

[Anonymous Monk]

Hi, the below script gets 2 sepereate hour and give me the difference between them. (ex, 08:00 - 17:00 = 9Hrs) the only problem is when i enter somthing like this( ex, (23:00 - 6:00(Am) = 17:00 should be 7) any idea how can i modify this to handle my problem or maybe use other module for me needs

#!/usr/bin/perl
use strict;
use warnings;
use Time::Local;

sub str2time{
       my $in = shift;
       my ($hour, $min, $sec) = split (/:/, $in);
       my ($mday,$mon,$year) = (localtime(time))[3,4,5];
       my $retn;
               eval {      $retn = timelocal($sec,$min,$hour,$mday,$mo
+n,$year);   };
        return $retn;
  }       
    
print "Please enter the first time in HH:MM:SS format: ";
my $intime = <STDIN>;
chomp $intime;
my $time1 = str2time($intime);
die "Invalid time: $intime\n" if !defined $time1;       

print "Please enter the 2nd   time in HH:MM:SS format: ";
$intime = <STDIN>;
chomp $intime;
my $time2 = str2time($intime);
die "Invalid time: $intime\n" if !defined $time1;

my $diff;
if ( $time1 > $time2 ) {    $diff = $time1 - $time2}
else {    $diff = $time2 - $time1}

my $mins  = int($diff / 60);
my $hours = int($mins / 60);
$mins     = $mins - ($hours * 60);
my $secs  = $diff - ($mins  * 60);
printf "Difference: %02d:%02d:%02d\n", $hours,$mins,$secs;

[Limbic~Region]

Anonymous Monk,
It appears that what you want to do is simple. If the LHS < RHS then RHS - LHS. If the LHS > RHS then (24:00 - LHS) + RHS. Obviously if LHS == RHS then 0. Note: LHS = left hand side and RHS = right hand side.

[si_lence]

You have to know if the first time is earlier than the second one. You could either ask for the date as well or assume that the first time entered is the earlier one and then handle your second case differently.

Without the date you will always have to make some assumptions about your input - it could be more than 24 hours between the two times ...

cheers

si_lence

[ikegami]

Without the date you will always have to make some assumptions about your input

Such as the lack to timezone change. For example, there can be 3 4, or 5 hours between 11pm and 3am depending on the date.

[Limbic~Region]

ikegami,
I really don't think Anonymous Monk is trying to do anything complicated. See Monthly hours report. I think this is just a matter of finding out how many hours between the first time entered and the second assuming you are always moving the clock forward (and not working for > 24 hrs consecutively). DST could still bite you but if you are writing your own code to manage your hours and haven't bothered to use a module - I say you get what you ask for.

[CountZero]

Assuming both times are in the same timezone and no change in DST between the two the anser is correct "+ k times 24 hours, for every integer value of k".

[ikegami]

Without date info, you can sort the times (if ( $time1 > $time2 )) or you can do what you asked, but you can't do both.

You problem can be solved by adding 24 hours to the later time, and modding the hours by 24.

use strict;
use warnings;

sub hms_to_s {
    my ($hms) = @_;
    my ($h,$m,$s) = split /:/, $hms;
    return $h*60*60 + $m*60 + ($s||0);
}

for (
    [qw( 08:00:00 17:00:00 )],
    [qw( 23:00:00 06:00:00 )],
) {
    my ($time1_hms, $time2_hms) = @$_;

    my $time1 = hms_to_s($time1_hms);
    my $time2 = hms_to_s($time2_hms);

    my $diff = ( ( $time2 + 24*60*60 ) - $time1 ) % (24*60*60);

    my $diff_h = int( $diff / (60*60) );
    my $diff_m = int( $diff / 60      ) % 60;
    my $diff_s =      $diff             % 60;

    printf("%s -> %s: %02d:%02d:%02d\n",
        $time1_hms,
        $time2_hms,
        $diff_h,
        $diff_m,
        $diff_s,
    );
}

08:00:00 -> 17:00:00: 09:00:00
23:00:00 -> 06:00:00: 07:00:00