use the non-greedy .+? modifier + the optional group (?: ... )*

my $test  = "Server list (Win2K/Win2003/XP) 6.26.5.1.5";
my $test1 = "Server list 6.26.5.1.5";

# $test =~ /(.*)(\(.+\))(.*)/;
my $regexp = qr{             # note: greedy (.+) vs. non-greedy (.+?) 
+qf.
                  (.+?)      # capture what comes but look ahead => $1
                  \s*        # optional whitespace
                  (?:          # start optional group (?: ... )?
                    \(         # opening bracket 
                    ([^)]+)    # get what's in the bracket => $2
                    \)         # closing bracket
                  )?           #  - optional
                  \s*        # possibly followed by some whitespace
                  ([\d.]+)   # then look for numbers and dots => $3
                  $          # which is anchored at the string end
               }x;
               
if( $test =~ /$regexp/ ) {  # if( $test1 =~ /$regexp/ ) 
   
   my ($prod_name, $platform, $version) = ($1, $2, $3);
   
   if(defined $platform) {
      print "$prod_name, $platform, $version"
   }
   else {
      print "$prod_name, <none>, $version"
   }
      
}
[download]

Regards

mwa

Thanks.Your solution is a perfect fit.

So you have to make them optional. But that menas that the the first .* will suck up all of the regex, so you have to find a compromise. The one I chose below is to allow everything but digits and opening parenthesis in the first chunk:

#!/usr/bin/perl
use strict;
use warnings;
my @tests = ("Server list (Win2K/Win2003/XP) 6.26.5.1.5", "Server list
+ 6.26.5.1.5");

for (@tests){
    if (m/([^(\d]+) ((?:\([^)]+\))?) (.*)/x){
        print "$1|$2|$3\n"
    } else {
        print "No match for $_\n";
    }
}
__END__
Server list |(Win2K/Win2003/XP)| 6.26.5.1.5
Server list ||6.26.5.1.5
[download]

You could use split, that splits a string into an array using a regex to determine the split position(s), as follow:

@result=split(/\s+\(|\)\s+/, $string);

This should split the string at any open parenthesis preceded by one or more spaces, and at any closed parenthesis followed by one or more spaces.

Hi psini,

That solution seems to be not fit here. In the 'test1' string, there is no parantheses to split, so it will obviously fail and you can't split the string.

OP wants to match the 'test1' string where the platform is missing and which is present in 'test' string.

Prasad

Yep, I realized it reading the other answers, sorry

Rule One: Do not act incautiously when confronting a little bald wrinkly smiling man.

Actually, IFF (if and only if) there are only the 2 variations presented in the OP, you could just use a

@arr=split(/ /,$test);

... and count the size of the array

$size=scalar(@arr);

if its 4, arr[2]is the OS, otherwise no OS was specified