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

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

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Or simply get what you need like this:

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

Online Demo

Answer 2

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

Answer 3

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.