I'm trying to list all users' home directories from a linux host.

You don't need a regex for that, you can get it directly from /etc/passwd. eg:

perl -F: -nae 'print "$F[0] has home dir $F[5]\n" if $F[5];' /etc/pass
+wd
[download]

Of course, you can change the test in there to be a regex match if you want to restrict the output, but it might be simpler to restrict it by UID range or similar. This should get you started anyway.

> you can get it directly from /etc/passwd

Unless the system uses LDAP :-)

($q=q:Sq=~/;[c](.)(.)/;chr(-||-|5+lengthSq)`"S|oS2"`map{chr |+ord
}map{substrSq`S_+|`|}3E|-|`7**2-3:)=~y+S|`+$1,++print+eval$q,q,a,
[download]

Be aware that Perl has a number of built-in functions relating to users and groups, for example: getlogin, getpwnam, getgrent. See Perl functions for fetching user and group info.

The simple code below, while perhaps not meeting your specific requirements, lists all users on a Linux box (including their home directories and other properties) by using the Perl builtin getpwent function:

use strict;
use warnings;

while ( my ($name, $passwd, $uid, $gid, $quota, $comment, $gcos,
            $home, $shell) = getpwent ) {
    print <<"ALL_USER_ATTRS";
name    = $name
passwd  = $passwd
uid     = $uid
gid     = $gid
quota   = $quota
comment = $comment
gcos    = $gcos
home    = $home
shell   = $shell

ALL_USER_ATTRS
}
[download]

See also Re: adding a group to the Unix System for example code that calls getpwnam.

G'day valerydolce,

Here's the steps you need for each line.

1. Match the start of the line.
2. Match the first / and everything up to, but not including, the second /.
3. Match a / followed by DUMMIES (one or zero times with no backtracking).
4. Match a /.
5. Start capture.
6. Match anything that isn't a / (one or more times non-greedily).
7. End capture.
8. Match the end of the line (before newline).

If the match was successful, $1 will contain 'user1', 'Dummy1', etc.

There a number of ways to implement this. I'll leave you to choose whichever you want.

Information on how to do everything described above can be found in perlre.

Update: I made the final step a little more specific by adding "(before newline)". Also fixed a typo: s/with contain/will contain/.

Hi valerydolce,

There are better solutions than the following (e.g. eyepopslikeamosquito's post), but since nobody has mentioned it yet, for completeness here are two solutions that actually access the file system and list directories. One of the problems with this is that there may be stray directories present that do not map to users; I've also hardcoded the "DUMMIES" name. In the second solution, I'm also showing something that I would normally recommend against: using a regex on a pathname, in this case it might be justified if you know you'll only be working with *NIX pathnames.

use Path::Class qw/file dir/;
my @homes = (
    grep({$_->is_dir && $_->basename ne 'DUMMIES'} dir('/home')->child
+ren),
    grep({$_->is_dir} dir('/home/DUMMIES')->children) );

use File::Spec::Functions qw/splitdir/; # core module
my @homes2 = (
    grep({-d && (splitdir($_))[-1] ne 'DUMMIES'} glob '/home/*'),
    #OR: grep({-d && !m#/DUMMIES$#} glob '/home/*'),
    grep({-d} glob '/home/DUMMIES/*') );
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Hope this helps,
-- Hauke D