Re: How do I replace certain character if condition exists

Possibly a simplistic solution, but have you tried using negative lookahead? For instance:

abowley@krait:~$ (echo 'Madrid(Spain' && echo 'Madrid(Spain)') | perl 
+-pe 's/\((?![^)]*\))/ /g;'
Madrid Spain
Madrid(Spain)
[download]

Breaking the regex down:

s/
  \(      # an open bracket
  (?!     # _not_ followed by
    [^)]* # anything except brackets (e.g. 'Spain')
    \)    # followed by a bracket
  )
 / /gz;
[download]

Comment on Re: How do I replace certain character if condition exists Select or Download Code

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Re: Re: How do I replace certain character if condition exists by Bilbo (Pilgrim) on Apr 17, 2003 at 13:41 UTC
This works if you are sure that you won't have any nested parentheses. If you do then it doesn't work at all, for example: a(b(c)) becomes a b(c)) a (b (c) d (e) f) becomes a b (c) d (e) f)	[reply]
Re: Re: Re: How do I replace certain character if condition exists by kilinrax (Deacon) on Apr 17, 2003 at 16:58 UTC
Well, you can do that with a regex, but it's a good degree more complicated..... `abowley@krait:~$ (echo 'Madrid(Spain(Europe)' && echo 'Madrid(Spain(Eu +rope))' && echo 'Madrid(Spain Europe))' && echo 'a(b(c))' && echo 'a +(b (c) d (e) f)') \| perl -mstrict -wpe 'BEGIN { $brackets = qr#$[^() +](?:(??{$brackets})[^()])*$# }; s#([^)]+?)($brackets)([^(]+)# ( my + $s = $1 ) =~ tr/(/ /; ( my $e = $3 ) =~ tr/)/ /; $s . $2 . $e #ge;' Madrid Spain(Europe) Madrid(Spain(Europe)) Madrid(Spain Europe) a(b(c)) a (b (c) d (e) f)` [download]	[reply] [d/l]


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