Adding to or Subtracting from a Date

Problem

You have a date and time and want to find the date and time of some period in the future or past.

Solution

Simply add or subtract Epoch seconds:

$when = $now + $difference;
$then = $now - $difference;

If you have distinct DMYHMS values, use the CPAN Date::Calc module. If you’re doing arithmetic with days only, use Add_Delta_Days ($offset is a positive or negative integral number of days):

use Date::Calc qw(Add_Delta_Days);
($y2, $m2, $d2) = Add_Delta_Days($y, $m, $d, $offset);

If you are concerned with hours, minutes, and seconds (in other words, times as well as dates), use Add_Delta_DHMS:

use Date::Calc qw(Add_Delta_DHMS);
($year2, $month2, $day2, $h2, $m2, $s2) = 
    Add_Delta_DHMS( $year, $month, $day, $hour, $minute, $second,
                $days_offset, $hour_offset, $minute_offset, $second_offset );

Discussion

Calculating with Epoch seconds is easiest, disregarding the effort to get dates and times into and out of Epoch seconds. This code shows how to calculate an offset (55 days, 2 hours, 17 minutes, and 5 seconds, in this case) from a given base date and time:

$birthtime = 96176750;                  # 18/Jan/1973, 3:45:50 am
$interval = 5 +                         # 5 seconds
            17 * 60 +                   # 17 minutes
            2  * 60 * 60 +              # 2 hours
            55 * 60 * 60 * 24;          # and 55 days
$then = $birthtime + $interval;
print "Then is ", scalar(localtime($then)), "\n";

                  Then is Wed Mar 14 06:02:55 1973

We could have used Date::Calc’s Add_Delta_DHMS function and avoided the conversion to and from Epoch seconds:

use Date::Calc ...

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