OK, sort of got somthing working but im sure there is a more "efficent" way to do it as its still returning a large array and loads of the elements are empty??? 
sub list{ 

my $path=shift;
my $comp=shift;
if (! -e $path){die "Error : $path $!\n";}
opendir(DIR,$path) or die "Error : $path $!\n";
return sort
 map {  my ($d,$m,$y) = (localtime( (stat "$path/$_")[9] ) )[3..5]; 
        $m+=1; $y+=1900;
        $m=($m<10)?"0$m":$m;
        $d=($d<10)?"0$d":$d;
        my $date = "$d/$m/$y";
        if($date eq $comp){"$_\n"};
    } grep(!/^(\.+?)$/,readdir(DIR));
}

[download]
any ideas??
Thanks for you help on this so far.
(goin to try the while() loop next)

In reply to Re^2: using grep on a directory to list files for a single date by markkneen
in thread using grep on a directory to list files for a single date by markkneen

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