my $str = 'xyzxyzxyz - Prod Cache Broker - 16100';
print "$1\n" if ($str =~ m/^(\S+)\s+?.+$/);
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The problem you were having is that (.+) is greedy, so it is consuming "xyzxyzxyz - Prod Cache Broker - 16100", and then the \s+? was forcing it to give up the space immediately preceding "16100", and finally the .+ was grabbing "16100".

If you wanted the parens to capture only up to (but not including) the first space, just tell it to match non-space characters.

See perlre for the non-greedy match operators, or maybe better, only capture the things you want to keep instead of capturing everything and later on figuring out what you don't want to keep.

How about using \w+ instead of .+ in your first capture?

$ perl -e 'my $str = "xyzxyzxyz - Prod Cache Broker - 16100";
print "$str\n$1\n" if $str =~ /(\w+)/;'
xyzxyzxyz - Prod Cache Broker - 16100
xyzxyzxyz
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Update: Disregard. Misread the OP. You say you do want only 'xyzxyzxyz'. Duh.

You say what you don't want, 'xyzxyzxyz', but don't say what you do want. I'm going to assume it's the phrase 'Prod Cache Broker'. Here's one approach. The definition of the  $word regex can be tuned to match the data you're handling.

c:\@Work\Perl\monks>perl -wMstrict -le
"my $str = 'xyzxyzxyz - Prod Cache Broker - 16100';
 ;;
 my $word = qr{ [[:alpha:]]+ }xms;
 ;;
 my ($phrase) = $str =~ m{ - \s+ ($word (?: \s+ $word)*) }xms;
 print qq{'$phrase'};
"
'Prod Cache Broker'
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