Perl alphabetic sort with alphanumeric string and regex - regex

Question

I am trying to sort alphanumeric string with perl and am facing the following problem:
Strings are like this: "XXXXX-1.0.0" where numbers can be on one or two digits and are representing release versions.
My problem is that when I have the two following strings:
- XXXXX-1.9.9
- XXXXX-1.10.0
XXXXX-1.9.9 is considered as greater than XXXXX-1.10.0 since 9 is greater than 1.
So I am trying to force numbers to be on two digits with a regexp.
There is the piece of code I am testing:
my $string = "XXXXXX-1.9.9";
my $pre = "0";
$string =~ s/(\.|-)+(\d{1})($|\.)/$1$pre$2$3/g;
print "$string\n";
This gives me the result "XXXXXX-01.9.09" which is not what I am looking for since it will not be sorted correctly. So I have to do that:
my $string = "XXXXXX-1.9.9";
my $pre = "0";
$string =~ s/(\.|-)+(\d{1})($|\.)/$1$pre$2$3/g;
$string =~ s/(\.|-)+(\d{1})($|\.)/$1$pre$2$3/g;
print "$string\n";
To get "XXXXXX-01.09.09"
My question is double:
- Is there a way to sort my strings with Perl without using regex?
- If I have to use regex is there a way to write it so that I do not have to execute it twice?
Thank you in advance.

Answer 1

You can use look around assertions to make sure there are not digits around the digit you are replacing without moving the position over the neighbouring characters.
#!/usr/bin/perl
use warnings;
use strict;
use feature qw{ say };
my #strings = qw( XXXXX-1.9.9
XXXXX-1.9.10
XXXXX-1.10.0 );
s/(?<=[^0-9]) ([0-9]) (?=[^0-9]|$) /0$1/xg for #strings;
#strings = sort #strings;
s/(?<=[^0-9]) 0+ ([0-9]) /$1/xg for #strings;
say for #strings;