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I'm building a profanity filter, and I'm using the following subroutine to replace matched words with XXXX:

while (($original, $converted) = @profanityArr) {
if (!$converted) {
$len = length($original);
$converted = "X" x $len;
}

$original = quotemeta($original);

$text =~ s/(\r|\n|\r\n|<br>|\s)*$original(\r|\n|\r\n|<br>|\s) */$1$converted$2/i;
}


# When I feed:
$original = "daym";

$text = "<br><br>daym<br><br>";
###

I'm getting "<br>XXXX<br>". Meaning, it loses the matched <br> in both $1 and $2.

# When I feed:
$original = "jason";
$converted = "brainfried";

$text = "<br><br>jason<br><br>";
###

I'm getting "<br>brainfried<br>". Again, it loses the matched <br> in both $1 and $2.


# When I feed:
$original = "dammit";
$converted = "XXXXit";

$text = "<br><br>dammit<br><br>";
###

I'm getting "<br>XXXXit<br><br>". Meaning, it loses the matched <br> in $1, but keeps it in $2.

It's the same if I change $1 and $2 to \1 and \2.

Any suggestions on how to correct the sub to keep the matched <br>?

"Jason C" <> wrote:

> $text =~
> s/(\r|\n|\r\n|<br>|\s)*$original(\r|\n|\r\n|<br>|\s) */$1$converted$2/i;
[...]
> # When I feed:
> $original = "daym";
>
> $text = "<br><br>daym<br><br>";
> ###
>
> I'm getting "<br>XXXX<br>". Meaning, it loses the matched <br> in both $1
> and $2.

You will want to capture the * also, otherwise $1 and $2 will
contain only one (the last) match for that part of the string:

((?:\r|\n|\r\n|<br>|\s)*)

The ?: will make the inner () non-capturing.

I think (but might be wrong - did not test) that \s contains
\r and \n, so you can remove them:

((?:<br>|\s)*)

Regards,
Jan

On Friday, June 22, 2012 1:37:51 AM UTC-4, Jan Pluntke wrote:
> You will want to capture the * also, otherwise $1 and $2 will
> contain only one (the last) match for that part of the string:
>
> ((?:\r|\n|\r\n|
> |\s)*)
>
> The ?: will make the inner () non-capturing.

Excellent! I was not familiar with the ?:, so I'll have to make a note of that for future reference.


> I think (but might be wrong - did not test) that \s contains
> \r and \n, so you can remove them:
>
> ((?:
> |\s)*)
>
> Regards,
> Jan

Correct again! I thought that \s just captured the space, and didn't realize that it includes line breaks (and apparently tabs, too). I can modify all of my scripts for that, now, and save a little bandwidth

Thanks for the help!

On Friday, June 22, 2012 5:07:40 AM UTC-4, Ben Morrow wrote:
> Unless $original is supposed to be a regex, you want \Q\E around it.

I originally did this in the function:

$original = quotemeta($original);
$text =~ ...;

Is there a difference between quotemeta() and \Q\E?


> You don't really need the final capture, you can just use lookahead.
> Similarly you don't need to capture more than one \s just to put it back
> again:
>
> s/(\s|
> ) \Q$original\E (?= \s|
> )/$1$converted/ix;
>
> Turning the initial capture into lookbehind is harder, since Perl
> doesn't support variable-length lookbehind and the two branches of the
> alternation are different lengths. However, if you have at least 5.10
> (which you do, I hope), you can use \K like this:
>
> s/ (?:\s|
> ) \K \Q$original\E (?=\s|
> ) /$converted/ix;

I'm afraid that you went just a little over my head on that one. What does the \K do? And what does (?=\s|<br>) do differently from (?:\s|<br>)? Or are they the same?

This is slightly different, but how do I include "or at the beginning of the string" in that regex?

I don't think that this would work, would it?

((?:^|\s|<br>)*)

For this purpose, I'm specifically converting a string of "www.example.com"to "http://www.example.com". A string like

$text = "Go to www.example.com";

matches, but

$text = "www.example.com<br><br>click here";

doesn't.

Further, in this case I don't want it to match when the www is between other characters (so that it doesn't change "http://www" to "http://http://www"), so I think I'll have to use a totally different regex without the trailing. But I still need to figure out how to make it match if it follows a \s,<br>, or is at the beginning of the string.

In article <s7pdb9->,
"Ben Morrow " <> spake thusly:
>
> s/ (?:\s|<br>) \K \Q$original\E (?=\s|<br>) /$converted/ix;

In addition to Jan and Ben's excellent suggestions, please note that
the patterns don't need to be applied in a loop. You can do something
like this:

my $regex=join "|", map quotemeta, @profanityArr;
$text =~ s{ (?:\s|<br>) \K ($regex) (?=\s|<br>) }{"X" x length $1}ex;

The "|" lets you match alternatives in a single regex, while the /e
flag is used to eval an expression before performing a substitution.

As I'm sure you know, folks who want to bypass the filters can usually
figure out ways around them.

- Morty