Try this.
This will retrun the difference in days between todays date and some other date.
my $start_date = "31-Oct-2000";
my $today = &date_to_unix(&get_date);
print "Content-type: text/plain\n\n";
$temp = int(((&date_to_unix($start_date) + ((&date_to_unix(&get_date)) * 86400)) - time())/86400);
$days = (($today) - ($temp));
print "Difference between start date and todays date in days - $days\n";
sub get_date {
# --------------------------------------------------------
# Returns the date in the format "dd-mmm-yyyy".
# Warning: If you change the default format, you must also modify the &date_to_unix
# subroutine below which converts your date format into a unix time in seconds for sorting
# purposes.
my ($time) = $_[0];
($time) or ($time = time());
my ($sec, $min, $hour, $day, $mon, $year, $dweek, $dyear, $daylight) = localtime($time);
my (@months) = qw!Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec!;
($day < 10) and ($day = "0$day");
$year = $year + 1900;
return "$day-$months[$mon]-$year";
}
sub date_to_unix {
# --------------------------------------------------------
# This routine must take your date format and return the time a la UNIX time().
# Some things to be careful about..
# int your values just in case to remove spaces, etc.
# catch the fatal error timelocal will generate if you have a bad date..
# don't forget that the month is indexed from 0!
#
my ($date) = $_[0];
my (%months) = ("Jan" => 0, "Feb" => 1, "Mar" => 2, "Apr" => 3, "May" => 4, "Jun" => 5,
"Jul" => 6, "Aug" => 7, "Sep" => 8, "Oct" => 9, "Nov" => 10,"Dec" => 11);
my ($time);
my ($day, $mon, $year) = split(/-/, $_[0]);
unless ($day and $mon and $year) { return undef; }
unless (defined($months{$mon})) { return undef; }
use Time::Local;
eval {
$day = int($day); $year = int($year) - 1900;
$time = timelocal(0,0,0,$day, $months{$mon}, $year);
};
if ($@) { return undef; } # Could return 0 if you want.
return ($time);
}
There may be a better way to do it, but I've tried this and it works.
Bob
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