Here's a simple trick, though perl, not PC: 

#!/usr/bin/perl
use File::Find;
use Getopt::Std;
use Cwd;
use strict;

my $u =  "usage: $0 -n count -s skip-pattern dir [dir ...]\n"
      . "  where count is number of newest files to report\n"
      . "  and skip-pattern is a comma separated list of regexes\n"
      . "  to apply upon files found, which will be - skipped\n";
die $u unless @ARGV;

my %o;
getopt('ns',\%o);

my $p; my $s;
if($o{s}) {
    ($p=$o{s}) =~ s/,/\|/g;
    $s++;
} 
my $n = $o{n} ? $o{n} - 1 : 10; # default 10 files
my @d = @ARGV;
push @d, getcwd unless @d;
my @f;
$#f = $n; # preallocate $n elements

! -d $_ and warn "directory $_: $!\n" and undef $_ for @d;
@d = grep { defined $_ } @d;
@d or die "no searchable directories in argument list\n";

find(\&wanted, @d);

print "skip-pattern: ($p)\n" if $o{s};
print join("\n",
    map { scalar(localtime $_->[0]).' '.$_->[1] } @f
),"\n";

sub wanted {
    my $file = $File::Find::name;
    -d && return;
    $s and  $file =~/($p)/ and return;
    my $time = (stat)[9] or die "can't stat $file: $!\n"; # mtime
    # if file is newer or as new than the first file...
    if ($f[0]->[0] < $time) {
        unshift @f,[$time,$file];
        pop @f if $#f > $n;
        return;
    } 
    # ...else insert the found file in the list.
    for( my $i = 0; $i<= $#f; $i++) {
        if($time >= $f[$i]->[0]) {
            die unless $time;
            splice @f,$i,0,[$time, $file];
            pop @f if $#f > $n;
            return;
        } 
    } 
}

[download]

update: improved code

In reply to Re: How to find the most recent file? by shmem
in thread How to find the most recent file? by zli034

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