Subtract months from given date - perl

Question

I have a date in the future, from which I have to subtract 3 months at a time until the current date is reached.
The reached date after subtracting the months must be closest to the current date, but has to be in the future.
The day of the month is always the 23rd
i.e.:
future date = 2015/01/23
current date = 2014/06/05
result = 2014/07/23
I'm running Solaris, so don't have access to GNU date.
I tried to do this in Perl, but unfortunately I can only use the Time::Local module:
#!/bin/ksh
m_date="2019/05/23"
m_year=$(echo $m_date|cut -d/ -f1)
m_month=$(echo $m_date|cut -d/ -f2)
m_day=$(echo $m_date|cut -d/ -f3)
export m_year m_month m_day
perl -MTime::Local -le '
$time = timelocal(localtime);
$i = 3;
while (timelocal(0, 0, 0, $ENV{'m_day'}, $ENV{'m_month'} - $i, $ENV{'m_year'}) > $time) {
print scalar(localtime(timelocal(0, 0, 0, $ENV{'m_day'}, $ENV{'m_month'} - $i, $ENV{'m_year'})));
$i += 3;
}
'
This only works for months within one year. Is there any other way I can do this?

Answer 1

It is simple enough to just split the date strings and do the arithmetic on the fields.
use strict;
use warnings;
use 5.010;
my $future = '2015/01/23';
my $current = do {
my #current = localtime;
$current[3] += 1;
$current[5] += 1900;
sprintf '%04d/%02d/%02d', #current[5,4,3];
};
my $result;
for (my $test = $future; $test gt $current; ) {
$result = $test;
my #test = split /\//, $test;
if (($test[1] -= 3) < 1) {
--$test[0];
$test[1] += 12;
}
$test = sprintf '%04d/%02d/%02d', #test;
}
say $result;
output
2014/07/23
Alternatively you could just do the division to calculate how many whole quarters to subtract, like this
use strict;
use warnings;
use 5.010;
my $future = '2015/01/23';
my #current = (localtime)[5,4,3];
$current[1] += 1;
$current[0] += 1900;
my #future = split /\//, $future;
my $months = ($future[0] - $current[0]) * 12 + $future[1] - $current[1];
$months -= 1 if $current[2] >= 23;
my #result = #current;
$result[2] = 23;
$result[1] += $months % 3;
$result[0] += int(($result[1] - 1) / 12);
$result[1] = ($result[1] - 1) % 12 + 1;
my $result = sprintf '%04d/%02d/%02d', #result;
say $result;
The output is identical to that of the previous code

Answer 2

This is your script changed so it should work across multiple years,
perl -MTime::Local -le'
sub nextm {
$ENV{m_year}--, $ENV{m_month} +=12 if ($ENV{m_month} -= 3) <1;
timelocal(0, 0, 0, $ENV{m_day}, $ENV{m_month}, $ENV{m_year});
}
my $time = timelocal(localtime);
while ((my $c=nextm()) > $time) {
print scalar localtime($c);
}
'

Answer 3

Try something like:
#!/usr/bin/perl -w
# just convert the real date that you have to epoch
my $torig = 1558569600;
my $tnow = time;
# 3 months in seconds to use the epoch everywhere
my $estep = 3 * 30 * 24 * 3600;
while(($torig - $estep) > $tnow){
$torig -= $estep;
}
print $torig,"\n";
print scalar localtime($torig),"\n";
The only problem here is that a months is an approximation, if you need the very same day but minus 3 months, you could use DateCalc

Answer 4

I ended up scripting it all in KSH instead of perl, thanks to Borodin's logic.
#!/bin/ksh
set -A c_date $(date '+%Y %m %d')
IFS=/ d="2019/05/23"
set -A m_date $d
[[ ${c_date[2]} -gt ${m_date[2]} ]] && ((c_date[1]+=1))
c_date[2]=${m_date[2]}
c_date[1]=$(( (((${m_date[0]} - ${c_date[0]}) * 12) + (${m_date[1]} - ${c_date[1]})) % 3 + ${c_date[1]} ))
if [[ ${c_date[1]} -gt 12 ]] ; then
((c_date[0]+=1))
((c_date[1]-=12))
fi
echo ${c_date[#]}