Convert Old Unix Date to Perl and compare - perl

Question

Requirement - I have file name called "Rajesh.1202242219". Numbers are nothing but a date "date '+%y''%m''%d''%H''%M'" format.
Now I am trying to write a perl script to extract the numbers from file name and compare with current system date and time and based on output of this comparison, print some value using perl.
Approach:
Extract the Digit from File name:
if ($file =~ /Rajesh.(\d+).*/) {
print $1;
}
Convert this time into readable time in perl
my $sec = 0; # Not Feeded
my $min = 19;
my $hour = 22;
my $day = 24;
my $mon = 02 - 1;
my $year = 2012 - 1900;
my $wday = 0; # Not Feeded
my $yday = 0; # Not Feeded
my $unixtime = mktime ($sec, $min, $hour, $day, $mon, $year, $wday, $yday);
print "$unixtime\n";
my $readable_time = localtime($unixtime);
print "$readable_time\n";
find Current time and compare...
my $CurrentTime = time();
my $Todaydate = localtime($startTime);
But the problem here is, I am not getting solution of how to extract 2 digit from $1 and assign to $sec, $min, etc. Any help?
Also, if you have good approach for this problem statement, Please share with me

Answer 1

I like to use time objects to simplify the logic. I use Time::Piece here because it is simple and light weight (and part of the core). DateTime can be another choice.
use Time::Piece;
my ( $datetime ) = $file =~ /(\d+)/;
my $t1 = Time::Piece->strptime( $datetime, '%y%m%d%H%M' );
my $t2 = localtime(); # equivalent to Time::Piece->new
# you can do date comparisons on the object
if ($t1 < $t2) {
# do something
print "[$t1] < [$t2]\n";
}

Answer 2

Might as well teach DateTime::Format::Strptime to make the comparison much simpler:
use DateTime qw();
use DateTime::Format::Strptime qw();
if (
DateTime::Format::Strptime
->new(pattern => '%y%m%d%H%M')
->parse_datetime('Rajesh.1202242219')
< DateTime->now
) {
say 'filename timestamp is earlier than now';
} else {
say 'filename timestamp is later than now';
};

Answer 3

my ($year, $month, $day, $hour, $min) = $file =~ /(\d{2})/g;
if ($min) {
$year += 100; # Assuming 2012 and not 1912
$month--;
# Do stuff
}

Answer 4

I think unpack might be a better fit.
if ( my ( $num ) = $file =~ /Rajesh.(\d+).*/ ) {
my ( $year, $mon, $day, $hour, $min ) = unpack( 'A2 A2 A2 A2 A2', $num );
my $ts = POSIX::mktime( 0, $min, $hour, $day, $mon - 1, $year + 100 );
...
}

Answer 5

Using a module that parses dates might be nice. This code will parse the date and return a DateTime object. Refer to the documentation to see the many ways to manipulate this object.
use DateTime::Format::Strptime;
my $date = "1202242219";
my $dt = get_obj($date);
sub get_obj {
my $date = shift;
my $strp = DateTime::Format::Strptime->new(
pattern => '%y%m%d%H%M'
);
return $strp->parse_datetime($date);
}