Need help creating a simple regex in Perl - regex

Question

I'm still useless when it comes to creating regex patterns. This one has got to be really easy.
Given the string: /home/users/cheeseconqueso/perl.pl
I want to place the string ./ right in front of the very last / in the original string to produce: /home/users/cheeseconqueso/./perl.pl
Thanks in advance - I'm sure this simple example will help me for a lot of other stupid stuff that shouldn't be up here!

Answer 1

Here's my solution based on what I was thinking of when I left the comment to your question:
use strict;
use warnings;
my $str = '/home/users/cheeseconqueso/perl.pl';
my #arr = split('/',$str);
my $newstr = join('/', #arr[0..(#arr-2)], '.', $arr[-1]);
Edit: If you're really keen on using a regex, this is the simplest one I've found:
$str =~ s|(.*/)(.*)|$1./$2|;
It takes advantage of the greediness of the initial * in the first group to take every character up to the last /, then matches everything else in the second group. It's a little easier to read with the | delimiters so you avoid leaning toothpick syndrome.

Answer 2

my $variable = "/home/users/cheeseconqueso/perl.pl";
$variable =~ s/(.*\/)([^\/]+)$/$1\.\/$2/;

Answer 3

this is using File::Basename not with regex, you can use this cpan module if you find hard to write regex.
use File::Basename;
$path = '/home/users/cheeseconqueso/perl.pl';
$newpath = dirname($path) . './'.basename($path);

Answer 4

You actually don't need a regex, you can just split the string:
my $str='/home/users/cheeseconqueso/perl.pl';
#Array: ('','home','users','cheeseconqueso','perl.pl')
my #arr=split(/\//,$str);
my $output=join('/',#arr[0..($#arr-2)]); # '/home/users/cheeseconqueso'
$output.='/./' . $arr[$#arr]; #Pops on the '/./' and the last element of #arr.

Answer 5

s/(.*)\/(.*)/\1.\/\2/
As noted by CanSpice, split works fine here too.

Answer 6

Try this:
my $str = "/home/users/cheeseconqueso/perl.pl";
if($str =~ /^(.*)(\/)(\w+\.\w+)$/) {
$str = $1.'/.'.$2.$3;
}

Answer 7

Regex? We don't need no stinkin' regex.
l-value substr() to the rescue!
#!/usr/bin/perl
use warnings;
use strict;
$_ = "/home/users/cheeseconqueso/perl.pl\n";
substr($_, rindex($_, '/'), 0) = '/.';
print;