How to extract directory names from a path in Perl - perl

Question

I have a path like this
/home/user/doc/loc
I want to extract home, user, doc, loc separately. I tried split (////) and also split("/")
but none of them worked. Please give me sample script:
while (<EXPORT>) {
if (/^di/) {
($key, $curdir) = split(/\t/);
printf "the current dir is %s\n", $curdir;
printf("---------------------------------\n");
($home_dir, $user_dir, $doc_dir, $loc_dir) = split("/");
}
}
But it didn't work; hence please help me.

Answer 1

Given $curdir containing a path, you'd probably use:
my(#names) = split m%/%, $curdir;
on a Unix-ish system. Or you would use File::Spec and splitdir. For example:
#!/usr/bin/env perl
use strict;
use warnings;
use File::Spec;
my $curdir = "/home/user/doc/loc";
my(#names) = split m%/%, $curdir;
foreach my $part (#names)
{
print "$part\n";
}
print "File::Spec->splitdir()\n";
my(#dirs) = File::Spec->splitdir($curdir);
foreach my $part (#dirs)
{
print "$part\n";
}
Ouput (includes a leading blank line):
home
user
doc
loc
File::Spec->splitdir()
home
user
doc
loc

Answer 2

split's first result will be the string preceding the first instance of the regular expression passed to it. Since you have a leading "/" here you would get an empty string in $home_dir, 'user' in $user_dir and so on. Add undef to the list assignment's first position or alternatively trim a leading slash first.
Also I'm not sure if you can call split without passing it $curdir here. Try:
(undef, $home_dir, $user_dir, $doc_dir, $loc_dir) = split("/", $curdir);