You can try this regex

if($test1 =~ m/(\S+)-(\S+)-([a-z]*)(\d+)-(\d\d)(\d+).*/)
{
    print $4,"|",$5;
}

I assume that u need only the first 2 didgits from 456562

perl -e '"xyz45sd2-32d34-sd23-456562.abc.com" =~ /(\d{2})-(\d{2})\d*(?=\.)/; print "$1\n$2\n"'

This other entry confirms that regex does not count: How to match word where count of characters same

Building upon GreatBigBore's idea, if there's an upper bound to the count, then you could try the or operator |. This only matches your requirement to find a match; depending on the matched count the match will be in different bins. Only one case correctly places them in $1 and $2. (\d{3})-(\d{3})|(\d{2})-(\d{2})|(\d{1})-(\d{1})

However if you concatenate the result captures as $1$3$5 and $2$4$6, you will effectively get the 2 stings you were looking for.

Another idea is to operate iteratively, you could repeat your search on the string by increasing the number until the match fails. (\d{1})-(\d{1}) , (\d{2})-(\d{2}) ...

A binary search comes to mind making it an O{ln(N)}, N being the upper limit for the capture length.

Theoretical answer

Short answer:

What you're looking for is not possible using regular expressions.

Long Answer:

Regular expressions (as their name suggests) are a compact representation of Regular languages (Type-3 grammars in the Chomsky Heirarchy).

What you're looking for is not possible using regular expressions as you're trying to write out an expression that maintains some kind of count (some contextual information other than beginning and end). This kind of behavior cannot be modelled as a DFA(actually any Finite Automaton). The informal proof of whether a language is regular is that there exists a DFA that accepts that language. As this kind of contextual information cannot be modeled in a DFA, thus by contradiction, you cannot write a regular expression for your problem.

Practical Solution

my ($lhs,$rhs) = $test =~ /^[^-]+-[^-]+-([^-]+)-([^-.]+)\S+/;
# Alernatively and faster
my (undef,undef,$lhs,$rhs) = split /-/, $test;

# Rest is common, no matter how $lhs and $rhs is extracted.
my @left = reverse split //, $lhs;
my @right = split //, $rhs;

my $i;
for($i=0; exists($left[$i]) and exists($right[$i]) and $left[$i] =~ /\d/ and $right[$i] =~ /\d/ ; ++$i){}

--$i;
$lhs= join "", reverse @left[0..$i];
$rhs= join "", @right[0..$i];

print $lhs, "\t", $rhs, "\n";