in reply to Re: Replace newlines only if not inside braces
in thread Replace newlines only if not inside braces

Works as long as }} is immediately followed by a \newline! ¹

Is there a reason why you've put the \K within the group?

I suppose this does the same and is better readable!

$data =~ s<(?:{{.*?}})?\K\n>{<br>\n}gs;
print $data;

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Cheers Rolf

1) see or clause in Re: Replace newlines only if not inside braces for a work around

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Re^3: Replace newlines only if not inside braces
by trizen (Hermit) on Feb 11, 2013 at 17:00 UTC
    No, there is no difference. I just thought it would be a little bit more efficient. I thought that, if the \K is outside, the $& variable is being cleaned up for every substitution, which is not really necessary. It should be cleaned only when something on the left side has been matched. Anyway, it is more readable in your way, and does, basically, the same thing. :)

    Alternatively, to work with strings that contain {{...}} groups, which are not followed by a newline, this code should do it: 
    $data =~ s<(?:{{.*?}}|[^\n])*\K\n>{<br>\n}gs;
print $data;

    [download]
[reply]
[d/l]
[select]
      $data =~ s<(?:{{.*?}}|[^\n])*\K\n>{<br>\n}gs;

      I think (didn't test it) that the * is two greedy, better *?.

      Cheers Rolf

[reply]
[d/l]
        Nope. It doesn't matter. The group doesn't match a single newline, so it can be greedy. It will try to match either the brackets block or any other character which is not a newline. If it succeeds, then tries again to match the block or a non-newline character. If it fails, it will match the newline character (\n) and makes the replace.

        Using the non-greedy quantifier, will do the same thing, but a little bit slower because it needs, unnecessary, to look ahead after each repetition (theoretically).
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[d/l]
Re^3: Replace newlines only if not inside braces
by 7stud (Deacon) on Feb 11, 2013 at 16:54 UTC

    Works

    Not quite.
[reply]

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