sub bydate{(stat $dir.$a)[9]<=>(stat $dir.$b)[9];}

@sortedfiles = sort bydate @files_in_your_dir;
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When I try to understand it, my brain hurts, because that'll be grossly inefficient (and perhaps even wrong if things change out from under you) for a dozen files or more. Consider the Schwartzian Transform instead.

-- Randal L. Schwartz, Perl hacker

#!/usr/local/bin/perl
use strict; 
$|++;

my $dir = "../../logs/";
my $output = "../stats/current.txt";

#read the dir and put all files to a list
opendir DIR, $dir || die "blah!";
my @allfiles = readdir DIR;
close DIR;

# remove those files from list that don't seem to be logfiles
my @logs = grep /^\d{6}.?\.log$/i, @allfiles;

# sort in a way you like
my @sorted = map{ $_->[0] }
             sort {$a->[1] cmp $b->[1]}
             map { [ substr($_,2,2).substr($_,0,2).substr($_,4)] }
             @logs;

# start output
open OUTPUT, ">$output" || die "blah!";
print OUTPUT join "\n", @logs;
close OUTPUT;
# end output
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