Regex for linkifying urls

Question

I'm trying to linkify URLs inside a text content. I know there are so many questions and answers for this purpose but I have a little bit different situation here.

What I want to do is converting this:

the best search engine is [out: http://google.com google].

into this:

the best search engine is <a href="http://google.com" rel="nofollow">google</a>.

or converting this:

the best search engine is [in: google].

into this:

the best search engine is <a href="http://mywebsite.com/google">google</a>.

What is the easiest way to do this in PHP for a newbie?

The best point I've reached is this:

$message = preg_replace("'(in: (.*))'Ui","(in: <a href=\"link.php?t=\\1\"><b>\\1</b></a>)",$message);

Please do not dump code in comments. Edit your original post to add any new information. — Jay Blanchard
– Jay Blanchard, Commented Jan 30, 2017 at 22:09
Why the difference between out: and in:? Why is that prefix necessary? — tadman
– tadman, Commented Jan 30, 2017 at 22:31
Actually, you don't need to use http://mywebsite.com/google, just /google will behave identically since it will look for an absolute path in the current website — Washington Guedes
– Washington Guedes, Commented Jan 30, 2017 at 22:45
@tadman just for better understanding and also to avoid converting every text inside brackets into links. — Sercan
– Sercan, Commented Jan 30, 2017 at 22:56

Nicolas · Accepted Answer · 2017-01-30 22:27:14Z

1

For the first one, you can use this regex: \[out:\s*([^\s]*)\s*(.*)\] and replace with <a href="\1" rel="nofollow">\2</a>.
For the second case, use \[in:\s*(.*)\], and replace with <a href="http://mywebsite.com/\1">\1</a>.

Here's some code to do it, using $result = preg_replace(pattern, substitution, input):

$result1 = preg_replace("/[out:\s*([^\s]*)\s*(.*)]/", '<a href="\1" rel="nofollow">\2</a>', $input);    
$result2 = preg_replace("/[in:\s*(.*)]/", '<a href="mywebsite.com\\1">\\1</a>', $input);

edited Jan 30, 2017 at 22:27

answered Jan 30, 2017 at 22:05

Nicolas

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4 Comments

Sercan Over a year ago

Thanks for the answer! I am able to run it after a few changes. Let me accept your answer after you have edit the code as follows for others:

Sercan Over a year ago

For the first one: $result = preg_replace("/[out:\s*([^\s]*)\s*(.*)]/", '<a href="\1" rel="nofollow">\2</a>', $input);

Sercan Over a year ago

For the second one: $result = preg_replace("/[in:\s*(.*)]/", '<a href="mywebsite.com\\1">\\1</a>', $input);

Nicolas Over a year ago

Was it the quotes in the pattern that were missing? I updated my answer.

ssc-hrep3 · Accepted Answer · 2017-01-30 22:38:31Z

You can use those two regular expressions:

\[in:\s*([^\[\]]*?)\]
\[out:\s*([^\[\]]*?)\s([^\[\]]*?)\]

Here is an example for the matching the groups in JavaScript (live test):

var regex1 = /\[in:\s*([^\[\]]*?)\]/g;
var regex2 = /\[out:\s*([^\[\]]*?)\s([^\[\]]*?)\]/g
var text = document.getElementById('main').innerHTML;
text = text.replace(regex1, '<a href="http://mywebsite.com/$1">$1</a>');
text = text.replace(regex2, '<a href="$1" rel="nofollow">$2</a>');

console.log(text);

<div id="main">
  the best search engine is [out: http://google.com google].
  the best search engine is [in: google].
</div>

And here the same in PHP:

<?php
$regex1 = "/\\[in:\\s*([^\\[\\]]*?)\\]/";
$regex2 = "/\\[out:\\s*([^\\[\\]]*?)\\s([^\\[\\]]*?)\\]/";
$text = 'the best search engine is [out: http://google.com google]. the best search engine is [in: google].';

$text = preg_replace($regex1, '<a href="http://mywebsite.com/$1">$1</a>', $text);
$text = preg_replace($regex2, '<a href="$1" rel="nofollow">$2</a>', $text);

echo $text;
?>

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