CODEWITHSUNDEEP
perl
Tree
Tree

Reputation: 10362

How can I separate a full path into directory and filename?

$a = '/etc/init/tree/errrocodr/a.txt'

I want to extract /etc/init/tree/errrocodr/ to $dir and a.txt to $file. How can I do that?

(Editor's note: the original question presumed that you needed a regular expression for that.)

Upvotes: 15

Views: 20686

Answers (7)

Tenuo
Tenuo

Reputation: 11

try this; it works at least for the filename, but when you modify it, it also gives you the direction: You can also modify it on UNIX-systems for the \ instead if / or to use them both with |

$filename =~ s/(^.*\/)//g;

Upvotes: 1

paxdiablo
paxdiablo

Reputation: 882346

Just use Basename:

use File::Basename;
$fullspec = "/etc/init/tree/errrocodr/a.txt";

my($file, $dir, $ext) = fileparse($fullspec);
print "Directory: " . $dir . "\n";
print "File:      " . $file . "\n";
print "Suffix:    " . $ext . "\n\n";

my($file, $dir, $ext) = fileparse($fullspec, qr/\.[^.]*/);
print "Directory: " . $dir . "\n";
print "File:      " . $file . "\n";
print "Suffix:    " . $ext . "\n";

You can see this returning the results you requested but it's also capable of capturing the extensions as well (in the latter section above):

Directory: /etc/init/tree/errrocodr/
File:      a.txt
Suffix:

Directory: /etc/init/tree/errrocodr/
File:      a
Suffix:    .txt

Upvotes: 25

living on edge
living on edge

Reputation: 471

Use @array = split("/", $a); and then $array[-1] is your file name.

Upvotes: 1

tryingToBeClever
tryingToBeClever

Reputation: 133

I think a regex solution is a perfectly legitimate need - since my googling for exactly that brought me here. I want to pull out a filename in a group that's part of a larger match expression - here's my attempt:

~> echo "/a/b/c/d" | perl -ne '/(?<dir>\/(\w+\/)*)(?<file>\w+)/ && print "dir $+{dir}  file $+{file}\n";'
dir /a/b/c/  file d

Upvotes: 1

ennuikiller
ennuikiller

Reputation: 46985

you don't need a regex for this, you can use dirname():

use File::Basename;
my $dir = dirname($a)

however this regex will work:

my $dir = $a
$dir =~ s/(.*)\/.*$/$1/

Upvotes: 16

leonbloy
leonbloy

Reputation: 76006

For example:

    $a =~ m#^(.*?)([^/]*)$#;
   ($dir,$file)  = ($1,$2);

But, as other said, it's better to just use Basename for this.

And, BTW, better avoid $a and $b as variable names, as they have a special meaning, for sort function.

Upvotes: 0

Piotr M&#252;ller
Piotr M&#252;ller

Reputation: 5558

Maybe this will work:

@^(.+/)/([^/]+)$@

Upvotes: -2

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