missing last character in perl regex

Question

(log doggies) (log needs)

^\(log (.*)[^)]\)\s*\(log (.*)[^)]\)$

It works with the exception of missing character at the end "s" as:

doggie need

Iterative coding assistance? http://stackoverflow.com/questions/32055326/how-to-capture-text-within-a-negate-class-character-using-perl http://stackoverflow.com/questions/31949142/whats-the-regular-expression-to-find-two-sets-of-parenthesis-in-a-row-using-per May I direct you to http://stackoverflow.com/q/22937618 to learn more about regex? — dlamblin, Aug 17 '15 at 17:22

score 0 · Answer 1 · answered Aug 17 '15 at 14:55

0

^\(log (.*)\)\s*\(log (.*)\)$

You don't need to negate the ).

answered Aug 17 '15 at 14:55

Takkun

score 0 · Answer 2 · answered Aug 17 '15 at 14:56

0

The [^)] eats your s-es. Why do you need it?

my $s = '(log doggies) (log needs)';
say for $s =~ /^\(log (.*)\)\s*\(log (.*)\)$/;

Output:

doggies
needs

answered Aug 17 '15 at 14:56

choroba

I need the [^)] to make sure it search for two consecutive sets of parenthesis and not three or four or .etc. – Bruce Walker Aug 17 '15 at 14:59
Do you need to use (.*)? If you're only concerned about alphanumeric characters you can use (\w*). – Takkun Aug 17 '15 at 15:13
@BruceWalker: Then please show the input that shows why you need it. – choroba Aug 17 '15 at 15:16

score 0 · Answer 3 · answered Aug 17 '15 at 17:35

It looks like your .* should be [^)]*, i.e. *any number of characters that are not closing parentheses. Giving

^\(log ([^)]*)\)\s*\(log ([^)]*)\)$

Or you could fetch all instances of (log xxx) with

while ( $s =~ /\(log ([^)]*)\)/g ) {
    print $1, "\n";
}

3 Answers3