You need to anchor your regex. Your regex is matching '!!' because it is matching an exclamation point followed by zero or more non exclamation-point characters anywhere in the string.
Thus the first '!' is matching the regex, and the second '!' is outside of your match. Try this: ################## use strict; use warnings; my $str = '!!'; print "$str\n" if $str =~ /^\![^!]*$/; ################# The ^ in the match denotes the start of the string, while the $ denotes the end (with an optional \n character). -----Original Message----- From: Jim [mailto:[EMAIL PROTECTED] Sent: Tuesday, April 18, 2006 5:52 PM To: beginners@perl.org Subject: Regex Help i am trying to match a '!' followed by any char but a '!' or no chars (string is only a '!') this is what I have and it is not working: $str = "!!"; # this is not working. it is matching "!!" print "$str\n" if $str =~ /\![^!]*/; Thanks for any help -- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] <http://learn.perl.org/> <http://learn.perl.org/first-response>
