in reply to Regex help

Not really sure what the point of this is, but if you add some more parens you should get what you want: 
my ($start,$middle,$end) = $string =~ /^(.*?)(($pattern)+)(.*?)$/g;

[download]
By placing the + outside of the first "parened" $pattern, you allow more than one - then, put some parens around that to catpure the results to $2.

Hope this helps, :)

UPDATE:
Oops, almost got that right ... now we are trying to match 4 items, not 3 anymore ... so try this: 
my ($start,$middle,undef,$end) = $string =~ ...

[download]
UPDATE 2:
I like liz's and CombatSquirrel's suggestion to use a look ahead non-capturing paren group ... but none of these solutions (yes bart, including mine ;)) are robust. For example, none will work with the pattern BABABABBB ...

UPDATE 3:
split! of course! i like it!

jeffa

L-LL-L--L-LL-L--L-LL-L--
-R--R-RR-R--R-RR-R--R-RR
B--B--B--B--B--B--B--B--
H---H---H---H---H---H---
(the triplet paradiddle with high-hat)

Replies are listed 'Best First'.
Re: Re: Regex help
by CombatSquirrel (Hermit) on Aug 24, 2003 at 15:39 UTC
    Errm, jeffa, that woud screw the my ($start,$middle,$end) part up.
    Suggested alternatives: 
    # first one
my ($start, $middle, undef, $end) = $string =~ /^(.*?)(($pattern)+)(.*
+?)$/g;

# second one, with non-capturing parens -- I like it better
my ($start,$middle,$end) = $string =~ /^(.*?)((?:$pattern)+)(.*?)$/g;

    [download]
    gri6507, note that /$pattern+/ is exactly the same as /AB+/ (well, at least for $pattern = 'AB'), which is, by definition /A(:?B)+/.
    Hope this helped.
    CombatSquirrel.

    Update: Arghh - wrong order for capturing and non-capturing parens. Fixed.

    Update 2: jeffa is right. The following RegEx should do the trick: 
    $pattern = 'AB';
'BABABABBB' =~
/
   ^                   # start at beginning of line
   (                   # capture to $1
      .*?                 # a number of character, but as few as possi
+ble ...
      (?<!$pattern)       # ... which may not contain $pattern
   )
   (                   # capture to $2
      (?:$pattern)+       # multiple occurences of $pattern
         |                   # OR
      (?!.*?$pattern)     # nothing, BUT there may be no $pattern in t
+he rest
                          #    of the string
   )
   (.*)                # capture rest to $3
/x ;
print "$1<$2>$3$/";
__END__
prints "B<ABABAB>BB"

    [download]
    I'm open for any suggestions, and yes, I do know Mastering Regular Expressions, I just forgot half (the important half) of it.

    Update 3 (Explanation): The RegEx engine tries to match at the earliest possible position. Therefore it will always match nothing to be captured in $1 (non-greedy dot-star), the highest possible number of following pattern matches (greedy star) and then the rest. Meaning, if the first pattern does not begin at the first character, $2 will also be empty (after all a star does not have to match) and the rest is slurped into $3. Bon appetit!
[reply]
[d/l]
[select]
Re: Re: Regex help
by BrowserUk (Patriarch) on Aug 24, 2003 at 15:44 UTC

    You can avoid the extraneous capture by using non-capturing parens. 

    my( $start, $middle, $end ) = $string =~ m[^ (.*?) ( (?:AB)+ ) (.*?) $
+]xg

    [download]

    ...but you know that:)

    Examine what is said, not who speaks.
    "Efficiency is intelligent laziness." -David Dunham
    "When I'm working on a problem, I never think about beauty. I think only how to solve the problem. But when I have finished, if the solution is not beautiful, I know it is wrong." -Richard Buckminster Fuller
    If I understand your problem, I can solve it! Of course, the same can be said for you.
[reply]
[d/l]
Re: Re: Regex help
by gri6507 (Deacon) on Aug 24, 2003 at 15:35 UTC
    Thank you, that does exactly what I need.
[reply]

In Section Seekers of Perl Wisdom