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.
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))