>>>>> "Sonika" == Sonika Sachdeva <[EMAIL PROTECTED]> writes:
>> You could instead keep in mind that: >> >> /^(?=.*foo)(?=.*bar)/ Ooops. that should be (?=.*?foo) for efficiency. >> my $regex = "^" . join "", map "(?=.*\Q$_\E)", @words; change .* to .*? there. -- Randal L. Schwartz - Stonehenge Consulting Services, Inc. - +1 503 777 0095 <merlyn@stonehenge.com> <URL:http://www.stonehenge.com/merlyn/> Perl/Unix/security consulting, Technical writing, Comedy, etc. etc. See PerlTraining.Stonehenge.com for onsite and open-enrollment Perl training! -- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] <http://learn.perl.org/> <http://learn.perl.org/first-response>
