如何在没有img src的情况下删除http和https

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.

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

Your pattern

/\b(https?):\/\/[-A-Z0-9+&@#\/%?=~_|$!:,.;]*[A-Z0-9+&@#\/%=~_|$]/i

removes all URLs (in the string) which start with protocol http or https. So when you apply it on your string, it will remove both that URL which is in the beginning of the string and that URL which is as src of <img>. So you have to use ^ in the beginning of your pattern:

$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;

Online Demo

Or simply get what you need like this:

/(<img.*\/>)/i

Online Demo

Try:

<[^>]*(*SKIP)(*FAIL)|\b(https?):\/\/[-A-Z0-9+&@#\/%?=~_|$!:,.;]*[A-Z0-9+&@#\/%=~_|$]

The <[^>]* catches all the things within an unclosed < and (*SKIP)(*FAIL)| skips them.

The rest is your regex.