use strict;
use warnings;
use DateTime;

my $now = DateTime->now(time_zone=>'local');
$now->truncate(to=>'day');

my $then = DateTime->new(time_zone=>'local',
    year=>2019, month=>11, day=>23 );

print $now->ymd," to ",$then->ymd,"\n";

my $diff = $now->delta_md($then);
my ($months,$days) = $diff->in_units('months','days');
print "$months months and $days days\n";

my $weeks = $now->delta_days($then)->in_units('weeks');
print "$weeks weeks\n";

__END__

2017-10-08 to 2019-11-23
25 months and 15 days
110 weeks
[download]

Worked like a champ! Thank you.

Here's yet another way to do it, using DateTime::Set for recurring events. It's also possible to modify the "recurrence" callback to shift the exact day if, for example, paydays can't be on a weekend, and there are even modules that calculate holidays for different regions.

use warnings;
use strict;
use DateTime;
use DateTime::Set;

my $paydays = DateTime::Set->from_recurrence(
    start  => DateTime->now,
    before => DateTime->new( year=>2019, month=>7, day=>1 ),
    recurrence => sub {
        my $dt = shift;
        return $dt->day < 15
            ? $dt->truncate(to=>'month')->add(days=>14 )
            : $dt->truncate(to=>'month')->add(months=>1)
    },
);

print "paydays: ",$paydays->count,"\n";
# - or -
my $iter = $paydays->iterator;
while ( my $dt = $iter->next ) {
    print $dt->ymd, "\n";
}

__END__

paydays: 41
2017-10-15
2017-11-01
2017-11-15
...
2019-05-15
2019-06-01
2019-06-15
[download]

Moreover, it's OK to post just a link to reddit.

($q=q:Sq=~/;[c](.)(.)/;chr(-||-|5+lengthSq)`"S|oS2"`map{chr |+ord
}map{substrSq`S_+|`|}3E|-|`7**2-3:)=~y+S|`+$1,++print+eval$q,q,a,
[download]

Ummmm...."Caution, Will Robinson!"

Perhaps your employer's procedure makes pay periods equivalent to the intervals between paydays... in which case, some paydays are for a longer pay period than others.

BUT
that just bends your mind and leads to complications beyond what you're trying to calculate.

All you really need to do -- based on the assertions in your OP -- is count months till you retire and multiply by two...and should you wish to use perl to do the arithmetic, you already have pointers to how to do it.

Hello Anonymous Monk,

Something "similar" has been asked before How to find business days?.

But also to point out that there is a beautiful module Date::Manip that contains a function called Date::Manip::Recur. There also some examples provided Date::Manip::Examples/RECURRING EVENTS which can retrieve the frequency from until specified date.

The beauty of this module is that you can change days, months, years, timezone(s) more or less what ever comes to your mind and you will get what you want.

Update For better and more accurate calculation use it like this:

my $start = ParseDate('now');
my $end = "2017123123:59:59";

# Instead of
my ($start, $end) = qw(2017-01-01 2017-12-31);
[download]

Sample of code:

#!/usr/bin/perl
use strict;
use warnings;
use Date::Manip;
use feature 'say';

my ($start, $end) = qw(2017-01-01 2017-12-31);

# y:m:w:d:h:m:s
my @days = ParseRecur('0:0:0:1:0:0:0', $start, $start, $end);
my @weeks = ParseRecur('0:0:1::0:0:0', $start, $start, $end);
my @months = ParseRecur('0:1:0::0:0:0', $start, $start, $end);
my @years = ParseRecur('1:0:0::0:0:0', $start, $start, $end);
my @datesEvery15th = ParseRecur('0:0:0:15:0:0:0', $start, $start, $end
+);

say "Number of days: " . scalar @days;
say "Number of weeks: " . scalar @weeks;
say "Number of months: " . scalar @months;
say "Number of years: " . scalar @years;
say "Number of payments: " . scalar @datesEvery15th;

=print payment dates
for my $date (@dates15th) {
    say UnixDate($date, "%Y-%m-%d");
}
=cut

__END__

$ perl test.pl
Number of days: 365
Number of weeks: 53
Number of months: 12
Number of years: 1
Number of payments: 25
[download]

Update2: To find out if the payment day is a working day or it will be shifted to the next working day. Read the POD for more information Date::Manip::DM5:

for my $date (@datesEvery15th) {
    if (Date_IsWorkDay($date)) {
    say UnixDate($date, "%Y-%m-%d");
    }
    else {
    my $nearestWorkingData = Date_NearestWorkDay($date);
    say UnixDate($nearestWorkingData, "%Y-%m-%d");
    }
}

__END__

2017-10-09
2017-10-24
2017-11-08
2017-11-23
2017-12-08
2017-12-22
[download]

Update3: An alternative way to know the exact time from today to what ever day you want that can easily be updated based on timezone. Sample of code below:

#!/usr/bin/perl
use strict;
use warnings;
use Date::Manip;
use feature 'say';
use Data::Dumper;

my $dateLocal = ParseDate('now');
# say $dateLocal;

my $end = "2017123123:59:59";

my $dateStartLocal = ParseDate($dateLocal);
my $dateEnd = ParseDate($end);

my $deltaLocal = DateCalc( $dateStartLocal,
               $dateEnd,
               1 );
say $deltaLocal;

my @dataLocal = split(/:/, $deltaLocal);

my @values = qw(year(s) month(s) week(s) day(s) hour(s) minute(s));

my %hashLocal;
@hashLocal{@values} = @dataLocal;

print Dumper \%hashLocal;

###### Different Timezone(s) ######

# From timeZone To timeZone
my $dateTimeZone = Date_ConvTZ($dateLocal,"GMT","CST");

my $dateStartTimezone = ParseDate($dateLocal);

my $deltaTimezone = DateCalc( $dateStartTimezone,
                  $dateEnd,
                  1 );
say $deltaTimezone;

my @dataTimezone = split(/:/, $deltaTimezone);

my %hashTinezone;
@hashTinezone{@values} = @dataTimezone;

print Dumper \%hashTinezone;

__END__

$ perl test.pl
0:2:3:1:9:13:31
$VAR1 = {
          'week(s)' => '3',
          'month(s)' => '2',
          'minute(s)' => '13',
          'hour(s)' => '9',
          'day(s)' => '1',
          'year(s)' => '0'
        };
0:2:3:1:9:13:31
$VAR1 = {
          'year(s)' => '0',
          'day(s)' => '1',
          'minute(s)' => '13',
          'hour(s)' => '9',
          'week(s)' => '3',
          'month(s)' => '2'
        };
[download]

Hope this helps, BR.

53 weeks in a year?



The way forward always starts with a minimal test.

Yes. The year is 365-366 days, depending, whereas weeks are 7 days (where 52 weeks is only 364 days: not a full year). Businesses often use "work weeks", and every once in a while, there is a "leap-week", to get the calendar back in sync, so the first work week of a year re-aligns with the start of the business's fiscal or calendar year. See more at ISO_week_date and ISO_8601.

Yup, if the year starts on a thursday (if you use ISO reckoning).

365 days are 52 weeks + 1 day. Every 6 years or so, you have 7 of these remaining days, and they, too, belong to a certain year…

Hello 1nickt,

You are right the year has 52 weeks and 1 or 2 day(s) depending upon the number of days in the year.

Sample of calculation on calendar common year 365 days:

1 common year = 365 days = (365 days) / (7 days/week) = 52.143 weeks =
+ 52 weeks + 1 day
[download]

Sample of calculation on leap year has 366 days, when February has 29 days:

1 leap year = 366 days = (366 days) / (7 days/week) = 52.286 weeks = 5
+2 weeks + 2 days
[download]

I was counting the remaining days as a week. So the solution is to remove the last input and count in total the remaining weeks:

Sample bellow before and after:

#!/usr/bin/perl
use strict;
use warnings;
use Date::Manip;
use Data::Dumper;

my ($start, $end) = qw(2017-01-01 2017-12-31);

# y:m:w:d:h:m:s
my @weeks = ParseRecur('0:0:1::0:0:0', $start, $start, $end);

print Dumper \@weeks;

=sample
$VAR1 = [
          '2017010100:00:00',
          '2017010800:00:00',
          '2017011500:00:00',
          '2017012200:00:00',
.
.
.
          '2017121700:00:00',
          '2017122400:00:00',
          '2017123100:00:00' # Not a week (remaining days)
        ];
=cut

pop @weeks;

say "Number of weeks: " . scalar @weeks;

print Dumper \@weeks;

=output
$VAR1 = [
          '2017010100:00:00',
          '2017010800:00:00',
          '2017011500:00:00',
          '2017012200:00:00',
.
.
.
          '2017121000:00:00',
          '2017121700:00:00',
          '2017122400:00:00' # Last week on 2017
        ];
=cut
[download]

BR / Thanos

Seeking for Perl wisdom...on the process of learning...not there...yet!