Use the file names as hash keys, and assign the age of the file to the hash value. Then sort the hash by value, and skip over the first ten. The rest you can blow away or otherwise do as you please. Something like this should get you started:

#! /usr/bin/perl -w

use strict;

my $directory = shift || '.';

opendir D, $directory or die "Cannot open directory $directory: $!\n";

my %age;
while( defined( my $file = readdir D )) {
    next if $file eq '.' or $file eq '..';
    my $canonical = "$directory/$file";
    $age{$canonical} = (stat $canonical)[9];
}
closedir D;

my $count = 0;
foreach my $file ( sort {$age{$b} <=> $age{$a}} keys %age ) {
    next if ++$count < 10;
    print "$file (@{[scalar localtime $age{$file}]})\n";
    # or unlink $file if you are feeling brave
}

[download]

update: Oops! I had the equality operator around the wrong way. As things stood, this code would have unlinked the ten most recent files and kept the rest! It's now around the right way. (Tip o' the hat to tomhukins). Also changed the code to reflect the sane comments of tachyon below.


--
g r i n d e r

In reply to Re: find "x" quantity newest files in a dir by grinder
in thread find "x" quantity newest files in a dir by braintoast

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