如何在没有img src的情况下删除http和https

Question

code:

$str = 'http://www.google.com <img src="http://placehold.it/350x150" />';
$str = preg_replace('/\b(https?):\/\/[-A-Z0-9+&@#\/%?=~_|$!:,.;]*[A-Z0-9+&@#\/%=~_|$]/i', '', $str);
echo $str;

output:

<img src="" />

i need this output:

<img src="http://placehold.it/350x150" />

how can i do it?

thanks for help.

Answer 1

I also think that DOMDocument and DOMXPath are preferable tools for parsing HTML markup.
But just in your particular case, here is solution with regexp negative lookbehind assertion :

$str = 'http://www.google.com <img src="http://placehold.it/350x150" /> http://www.google.com.ua';

$str = preg_replace('/(?<!src=\")(https|http):\/\/[^\s]+\b/i', '', $str);

print_r($str);   // <img src="http://placehold.it/350x150" />

This will remove all urls excepting those which are inside an img src attribute