[RESOLVED] Preg Replacing urls in strings with bbcode

Undrium

Heyas,

I have a quite simple problem I think, but I haven't really nailed it yet. I'm trying to replace all occurances of urls within a text with the bbcode elements.


//$indata = "testing this out http://www.google.com";
preg_match('/(http:\/\/[^\s]+)/', $indata, $text);
$hypertext = "[url=". $text[0] . "]" . $text[0] . "[/url]";
$indata = preg_replace('/(http:\/\/[^\s]+)/', $hypertext, $indata);

//indata is now "testing this out [url= http://www.google.com] http://www.google.com[/url]"

This works fine, until an input with bbcode occurs. Then all urls gets duplicate bbcode, looking like this instead:

[url=[url=http://www.google.com] http://www.google.com[/url]][url=http://www.google.com] http://www.google.com[/url][/url]"

There must be a solution to this out there, I just haven't found it? All help is extremely appreciated.

vicodin

You must be running a loop twice some where else in your code.

I did this:

$indata = "This is a Link http://www.google.com";
If (preg_match('/(http:\/\/[^\s]+)/', $indata, $text)){    

$hypertext = "[url=". $text[0] . "]" . $text[0] . "[/url]";
$indata = preg_replace('/(http:\/\/[^\s]+)/', $hypertext, $indata); 
}

if(empty($indata)){
echo 'Its Empty';
}Else{
    echo $indata;
}

And came out with this:

This is a Link {url=http://www.google.com]http://www.google.com{/url]

I had to change the first square brackets to curley brackets so you could see the exact output.

Undrium

Thanks for your reply, vicodin!

My problem is not that the code gets repeated, the problem occurs when the indata already contains bbcode and gets replaced again. If it already has been replaced with bbcode url-blocks, I do not want to touch that url.

johanafm

Use assertions so that your pattern disregards any url that has [url= in front of it.

Undrium

johanafm: Thanks for the hint, I yet have problem to figure out how preg works. Could you give me a little heads up how I should use assertions in this case?

johanafm

Using assertions is as easy as the docs say, as long as you remember that they can't have variable length. But since that isn't needed here anyway, there really is no problems at all

# negative look behind assertions look like this (read as: pattern not preceeded by)
$pattern = '(?<!';

# You don't want to capture when it's on the form
# 	[url=http://example.com/file.ext]link text[/url]
# but you do want to capture when it is simply
# 	[url]http://example.com/file.ext[/url]
# So, when pattern is not preceeded by [url= and [ is a special pattern character that needs escaping
$pattern = '(?<!\[url=)';

# Which using your original pattern
# 	/(http:\/\/[^\s]+)/
# would give
$pattern = '/(?<!\[url=)(http:\/\/[^\s]+)/';

And the rest is just feedback for improvement

# and then we change opening and closing delimiters from / so we don't have to escape those
$pattern = '#(?<!\[url=)(http://[^\s]+)#';

# and then we get rid of your parenthesis, which is a capturing subpattern, since you don't use
# it anyway - you used the whole pattern, which is the exact same thing when the entire pattern
# is also a capturing subpattern.
# do note that the first set of parenthesis is NOT a capturing subpattern. It is an assertion and 
# requires the parenthesis
$pattern = '#(?<!\[url=)http://[^\s]+#';

# And finally we get rid of your use of preg_match followed by preg_replace, since you can use $0
# in your replacement text to refer to the captured pattern when using preg_replace
$replace = '[url=$0]$0[/url]';

# and now you do all the work in one reg exp pass, instead of doing it in two as before
# And should you ever find yourself doing one _match followed by _replace, also capture
# the offset with match, PREG_OFFSET_CAPTURE (see docs for preg_match)
# and then use ordinary string replacement at that offset, since you now know exactly where
# the text is and what the text is, so you no longer need any kind of pattern matching.
$s = preg_replace($pattern, $replace, $search);

Undrium

Wow, johanafm, thanks alot for your detailed help (great effort!)! I am sure alot of others will find this information useful, you are a hero. 🙂

Undrium

Apparantly I took out the victory too early, when I did test this at home I got into the same problem as before. I think the dilemma is that

[url=http://www.google.com]http://www.google.com[/url]

Has two "http" where only the first one matches the condition of johans pattern. :/

johanafm

This might do the trick then

$s = "[url=http://www.google.com]http://www.google.com[/url]
[url]http://example.com[/url]
see (http://second.example.com)";

$pattern = '#(?<=\s|\()(http://[^\s\)]+)#';

Assert that the url is preceeded by a whitespace (\s) or an opening parenthesis,
as in "see (http://example.com)
Neither of these are true for [ url=...]htt...[ /url].

Undrium

Heyas, and thanks for stopping by again. 😉

The problem as I see it is that asserting whitespace won't work if the url is the first word in the string (since the start has no whitespace) also if it is a newline the url won't work.

I didn't find a good solution for this, what I did was checking if the http had a "]" (plus the code johan has provided) in front of it, it works but does not support:

[url]someurl[/url]

which is quite sad.

johanafm

Dealing with start of string is easy. All you do is insert the start of string, ^, into the lookbehind assertion, i.e. (?<=^|\s|(), so it checks that the pattern is preceeded by
^ or whitespace or (

I'm not skilled enough to construct one single pattern to replace both [ url]...[ /url], ... while skipping [ url=...]...[/url]. But two patterns does the trick without problems.

$patterns = array('#\[url](.+)\[/url]#', '#(?<=^|\s|\()(http://[^\s\)]+)#');
$s = preg_replace($patterns, '[url=$1]$1[/url]', $s);

Undrium

Cheers, Johan, that is a great solution. 🙂

laserlight

Incidentally, there is a bbcode extension that you may be able to use.