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Home/ Questions/Q 3981428
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Editorial Team
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Editorial Team
Asked: May 20, 20262026-05-20T05:24:31+00:00 2026-05-20T05:24:31+00:00

I have this code which lists all files in my directory: $dir = ‘/var/www/corpIDD/rawFile/’;

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I have this code which lists all files in my directory:

$dir = '/var/www/corpIDD/rawFile/';
opendir DIR, $dir or die "cannot open dir $dir: $!";
my @file= readdir DIR;
closedir DIR;

which returns an array containing something like this:

$array (0 => 'ipax3_2011_01_27.txt', 1 => 'ipax3_2011_02_01.txt', 2 => 'ipax3_2011_02_03.txt')

My problem here is, how will I store elements 1 => ‘ipax3_2011_02_01.txt’ and 2 => ‘ipax3_2011_02_03.txt’ to separate variable as they belong to the same month and year(2011_02)?

Thanks!

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-20T05:24:32+00:00Added an answer on May 20, 2026 at 5:24 am

    In Perl, when you need to use a string as the key in a data structure, you are looking for the HASH builtin type, designated by the % sigil. A nice feature of Perl’s hashes is that you do not have to pre-declare a complex data structure. You can use it, and Perl will infer the structure from that usage.

    my @file = qw(ipax3_2011_01_27.txt ipax3_2011_02_01.txt ipax3_2011_02_03.txt);
    
    my %ipax3;
    
    for (@file) {
        if (/^ipax3_(\d{4}_\d{2})_(\d{2}).txt$/) {
            $ipax3{$1}{$2} = $_
        }
        else {
            warn "bad file: $_\n"
        }
    }
    
    for my $year_month (keys %ipax3) {
        my $days = keys %{ $ipax3{$year_month} };
        if ($days > 1) {
            print "$year_month has $days files\n";
        }
        else {
            print "$year_month has 1 file\n";
        }
    }
    

    which prints:

    2011_01 has 1 file
    2011_02 has 2 files
    

    To get at the individual files:

    my $year_month = '2011_02';
    my $day        = '01';
    my $file       = $ipax3{$year_month}{$day};
    

    Above I used the return value of the keys function as both the list to iterate over, and as the number of days. This is possible because keys will return all of the keys when in list context, and will return the number of keys in scalar context. Context is provided by the surrounding code:

    my $number = keys %ipax3; # number of year_month entries
    
    my @keys = keys %ipax3; # contains ('2011_01', '2011_02') 
    
    my @days = keys %{ $ipax{$year_month} };
    

    In the last example, each value in %ipax is a reference to a hash. Since keys takes a literal hash, you need to wrap $ipax{$year_month} in %{ ... }. In perl v5.13.7+ you can omit the %{ ... } around arguments to keys and a few other data structure access functions.

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