Question

Is there an elegant way in Perl to find the newest file in a directory (newest by modification date)?

What I have so far is searching for the files I need, and for each one get it’s modification time, push into an array containing the filename, modification time, then sort it.

There must be a better way.

Answer 1

Your way is the ‘right’ way if you need a sorted list (and not just the first, see Brian’s answer for that). If you don’t fancy writing that code yourself, use this

use File::DirList; my @list = File::DirList::list('.', 'M'); 

Personally I wouldn’t go with the ls -t method – that involves forking another program and it’s not portable. Hardly what I’d call ‘elegant’!

---

Regarding rjray’s solution hand coded solution, I’d change it slightly:

opendir(my $DH, $DIR) or die 'Error opening $DIR: $!'; my @files = map { [ stat '$DIR/$_', $_ ] } grep(! /^\.\.?$/, readdir($DH)); closedir($DH);  sub rev_by_date { $b->[9] <=> $a->[9] } my @sorted_files = sort rev_by_date @files; 

After this, @sorted_files contains the sorted list, where the 0th element is the newest file, and each element itself contains a reference to the results of stat, with the filename itself in the last element:

my @newest = @{$sorted_files[0]}; my $name = pop(@newest); 

The advantage of this is that it’s easier to change the sorting method later, if desired.

---

EDIT: here’s an easier-to-read (but longer) version of the directory scan, which also ensures that only plain files are added to the listing:

my @files; opendir(my $DH, $DIR) or die 'Error opening $DIR: $!'; while (defined (my $file = readdir($DH))) {   my $path = $DIR . '/' . $file;   next unless (-f $path);           # ignore non-files - automatically does . and ..   push(@files, [ stat(_), $path ]); # re-uses the stat results from '-f' } closedir($DH); 

NB: the test for defined() on the result of readdir() is because a file called ‘0’ would cause the loop to fail if you only test for if (my $file = readdir($DH))