Multiple regex in function using php

Question

I wanted to pass data that contains url within it, and the url will be converted to click-able links, but the issue is that when i run the function, it reiterate 3 times and the data is treated 3 times. How can I have a single data output? I tried removing the concatenation .= to just = but then, only the last pattern is being treated. I want it in 3 pattern because I want to add http to the href when a user input is www.ivotism.com, it will be <a href="http//www.ivotism.com">www.ivotism.com</a> instead of <a href="www.ivotism.com">www.ivotism.com</a>.

function linkify($inputText) {
        //URLs starting with http://, https://, or ftp://
        $replacePattern1 = '/(\b(https?|ftp):\/\/[-A-Z0-9+&@#\/%?=~_|!:,.;]*[-A-Z0-9+&@#\/%=~_|])/i';
        $replacedText .=  preg_replace($replacePattern1, '<a href="$1" target="_blank">$1</a>', $inputText);

        //URLs starting with "www." (without // before it, or it'd re-link the ones done above).
        $replacePattern2 = '/(^|[^\/])(www\.[\S]+(\b|$))/i';
        $replacedText .=  preg_replace($replacePattern2, '$1<a href="http://$2" target="_blank">$2</a>', $inputText);

        //Change email addresses to mailto:: links.
        $replacePattern3 = '/(([a-zA-Z0-9\-\_\.])+@[a-zA-Z\_]+?(\.[a-zA-Z]{2,6})+)/i';
        $replacedText .=  preg_replace($replacePattern3, '<a href="mailto:$1">$1</a>', $inputText);

        return $replacedText;
    }

I run the code as shown below:

$ab = "<br>1 http://www.ivotism.com <br>2 https://www.ivotism.com/hom.php?u=kira&id=2 <br>3 ftp://www.ivotism.com <br>4 w www.ivotism.com <br>5 [email protected]"; echo 
linkify($ab);

This is the result I got:

1 http://www.ivotism.com
2 https://www.ivotism.com/hom.php?u=kira&id=2
3 ftp://www.ivotism.com
4 www.ivotism.com
5 [email protected]
1 http://www.ivotism.com
2 https://www.ivotism.com/hom.php?u=kira&id=2
3 ftp://www.ivotism.com
4 www.ivotism.com
5 [email protected]
1 http://www.ivotism.com
2 https://www.ivotism.com/hom.php?u=kira&id=2
3 ftp://www.ivotism.com
4 www.ivotism.com
5 [email protected]

You're using $inputText as the input each time. You need to use the result of the previous replacement. — Barmar
– Barmar, Commented Dec 14, 2014 at 13:30

Barmar · Accepted Answer · 2014-12-14 13:46:24Z

1

You can perform all the replacements in a single call to preg_replace, by putting the regular expressions and replacements into arrays.

$resultText = preg_replace(array($replacePattern1, $replacePattern2, $replacePattern3),
                           array($replaceSub1, $replaceSub2, $replaceSub3),
                           $inputText);

answered Dec 14, 2014 at 13:46

Barmar

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1 Comment

Mande Kira Over a year ago

Thats great, and there is no having problem with function redeclaration. its ok.

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