How can I identify an argument as a year in Perl?

Question

I have created a file argument.pl which takes several arguments first of which should be in form of a year For example: 2010 23 type. Here 2010 is a year

my code does something like:

use strict;
use warning
use Date::Calc qw(:all);
my ($startyear, $startmonth, $startday) = Today();
my $weekofyear = (Week_of_Year ($startyear,$startmonth,$startday))[0];
my $Year = $startyear;
...
...

if ($ARGV[0])
{
$Year = $ARGV[0];
}

Here this code fills $Year with "current year" if $ARGV[0] is null or doesn't exist. What do I use here instead of if ($ARGV[0])?

Is it possible to check that the value in $ARGV[0] is a valid year (like 2010, 1976,1999 etc.)?

Answer 1

You can try the following to ensure the first argument is of the format 19xx or 20xx.

if ($ARGV[0] =~/^(19|20)\d{2}$/) {
 $Year = $ARGV[0];
}

Answer 2

Simple test for defined-ness and 4-digits format:

$Year = $ARGV[0] if defined $ARGV[0] && $ARGV[0] =~ /^\d{4}$/);

Answer 3

give it a try, to find the year is 20** or 19**

if ($ARGV[0] =~ m!^((?:19|20)\d\d)) {
$year = $ARGV[0];
}

Hope this helps.

Answer 4

@OP, since you are using Date::Calc, looking at the documentation, you might want to try using some of the functions such as check_date() eg

check_date
      if (check_date($year,$month,$day))

if your year is not correct, the function should give you an error.

Answer 5

I would recommend ditching positional arguments and using a command line argument processing library such as Getopt::Long so that your program can be called as

$ argument.pl -y 2013 -d 23 -t type