[ACCEPTED]-Getting the list of subdirectories (only top level) in a directory using Perl-perl
Accepted answer
If you want to collect the dirs into an 2 array:
my @dirs = grep {-d "$root/$_" && ! /^\.{1,2}$/} readdir($dh);
If you really just want to print the 1 dirs, you can do:
print "$_\n" foreach grep {-d "$root/$_" && ! /^\.{1,2}$/} readdir($dh);
next unless $name =~ /^\.\.?+$/;
Also, the module File::Find::Rule makes 1 a vary nice interface for this type of thing.
use File::Find::Rule;
my @dirs = File::Find::Rule->new
->directory
->in($root)
->maxdepth(1)
->not(File::Find::Rule->new->name(qr/^\.\.?$/);
Just modify your check to see when $name 1 is equal to '.' or '..' and skip the entry.
File::Slurp read_dir automatically excludes the special dot 4 directories (. and ..)
for you. There 3 is no need for you to explicitly get rid 2 of them. It also performs checking on opening 1 your directory:
use warnings;
use strict;
use File::Slurp qw(read_dir);
my $root = 'mydirectoryname';
for my $dir (grep { -d "$root/$_" } read_dir($root)) {
print "$dir\n";
}
use warnings;
use strict;
my $root = "mydirectoryname";
opendir my $dh, $root
or die "$0: opendir: $!";
while (defined(my $name = readdir $dh)) {
next unless -d "$root/$name";
next if $file eq ".";
next if $file eq "..";
print "$name\n";
}
0
Source:
stackoverflow.com
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