Preg replace image src using callback

Question

Preg replace image src using callback

1.4k views Asked by Michak Firas At 12 September 2024 at 10:34

I have an article with HTML tags. It's a long article with 3/5 images. Now I want to update every image src attributes. Example:

Image html tag looks like:

<img class="aligncenter" style="display: block;margin-left:auto;margin-right:auto;" src="http://img.zszywka.com/0/0269/w_0980/moj-swiat/muza-2013-najnowsze-eska-hity-2013-.jpg" width="642" />

I want to take this URL, make some changes and next update src. Then go to next image and do it again (so script must change all images src)

Final img tag looks like:

<img class="aligncenter" style="display: block;margin-left:auto;margin-right:auto;" src="http://EXMAPLE.COM/0/0269/w_0980/moj-swiat/muza-2013-najnowsze-eska-hity-2013-.jpg" width="642" />

So I need to manipulate with changes. I try using preg_replace_callback but I have a problem with it:

// change image src


$finalContent = preg_replace_callback('/' . preg_quote('src="(*.?)"') . '/', 
function() use ($variable_with_changes){ return $variable_with_changes; }, $variable_with_article_content);

echo $finalContent;

This doesn't works, I don't have an idea how I can update image domain and keep path.

Original Q&A

There are 3 answers

chris85 On 16 December 2016 at 17:48

The parsing answer is better but to answer your question of why your regex/preg_replace_callback fails...

preg_quote escapes all special regex characters, so don't use the on your regex. If you were passing variables or unknown characters to your regex you'd use that.
*. is invalid the * is a quantifier, meaning 0 or more of the previous character/group are allowed. .* is what you wanted, it means zero or more of any character excluding new lines.
You are not doing anything with your found matches, you need to pass them into the function. function($match).
You never make a swap of domain names in your function. You can use str_replace for that.

PHP Example:

$variable_with_article_content = '<img class="aligncenter" style="display: block;margin-left:auto;margin-right:auto;" src="http://img.zszywka.com/0/0269/w_0980/moj-swiat/muza-2013-najnowsze-eska-hity-2013-.jpg" width="642" />';
$finalContent = preg_replace_callback('/' . 'src="(.*?)"' . '/', 
function($match) { return 'src="' . str_replace('img.zszywka.com', 'EXMAPLE.COM', $match[1]) . '"'; }, $variable_with_article_content);

echo $finalContent;

Demo: https://eval.in/699201

Note also your regex has no requirement that you are replacing an image's domain.

Michak Firas On 16 December 2016 at 17:57

I found solution, and works great. :)

function getURL($matches) {
  global $rootURL;
  return $matches[1] . $rootURL . "?type=image&URL=" . base64_encode($matches['2']);
}

$contents = preg_replace_callback("/(<img[^>]*src *= *[\"']?)([^\"']*)/i", getURL, $contents);

Thank's all for reply to me!

**miken32** · Accepted Answer · 2016-12-16 17:42:17

You should be parsing HTML as HTML, not using regular expressions.

$doc = new DOMDocument();
$doc->loadHTML('<html><body><img class="aligncenter" style="display: block;margin-left:auto;margin-right:auto;" src="http://img.zszywka.com/0/0269/w_0980/moj-swiat/muza-2013-najnowsze-eska-hity-2013-.jpg" width="642" /></body></html>');
$images = $doc->getElementsByTagName('img');
foreach ($images as $img) {
    $url = $img->getAttribute('src');
    // do whatever you need to with $url
    $url = str_replace('img.zszywka.com', 'example.com', $url);
    $img->setAttribute('src', $url);
}
echo $doc->saveHTML();

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Preg replace image src using callback

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