Re: Sorting dir output

[yaron]

Hi,

If you wish to select all files that are directly under given directory you can 
implement the following (in pure perl fashion):


sub numerically { $b <=> $a;}

$DIR = <Set in you directory>;

$THRESHOLD_IN_DAYS = 3;

my %time_to_file;
my $currTime = time();
#Store all file in hash with time as key.
for each my $file ( grep (-f,<*>)) { 
    my $time = stat($file)->mtime;
    $time_to_file{$currTime - $time} = [] unless ( exists 
$time_to_file{$currTime - $time});
    push @{$time_to_file{$currTime - $time}},$file;
}

#Sort by time from the new to the old
my @sorted = sort numerically keys(%time_to_job);

my $time_threshold_in_seconds = $THRESHOLD_IN_DAYS * 24 *3600

foreach my $atime (@sorted) {
    if ($atime >= $time_threshold_in_seconds) {
        foreach my $afile ($time_to_file{$atime}) {
            print "$afile is older than $THRESHOLD_IN_DAYS lets delete it\n";
            unlink($afile) or warn "Cannot delete $afile. $!";
        }
     }
}

Hope that helps

Yaron Kahanovitch



----- Original Message -----
From: "Craig Schneider" <[EMAIL PROTECTED]>
To: beginners@perl.org
Sent: Tuesday, April 10, 2007 2:19:40 PM (GMT+0200) Auto-Detected
Subject: Sorting dir output

Hi Guys

How could I exec a 'dir' command on a dos system and put the output in
an array, sort by date and the files that are older than 3 days be moved
into a folder called 'history'

Thanks

Craig

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