当src不在示例域上时，Preg替换所有img

Question

I have a problem with preg_replace, my function have some bug and i dont know where. I need to remove from array every image that has source other than my main domain.

This is my function:

$aPatterns = array (
(...),
'#<img[^>]+src="(?!http://example.com/emoticon/(example|emot|name).gif)[^\s]+"[^>]+>#is',
(...));

$aReplecements = array(
(...),
'',
(...)
);

$contentOut = preg_replace($aPatterns,$aReplecements,$content);

If img tag is not closed or does not have double quotes sign at the end of src attribute function not working, no change to empty string.

I have no idea what is wrong in this expression, I hope, and I count on your help.

Regards,

Answer 1

Okay, so the problem is that you have explicitly stated that there should be a " and > at the end of the URL (you've also explicitly stated that there should be an opening " at the beginning of the URL after the src= so if that wasn't there it wouldn't work either).

You can fix this behaviour by making the " and > optional by adding a ? mark after them.

#<img[^>]+src="?(?!http://example.com/emoticon/(example|emot|name).gif)[^\s]+"?[^>]+>?#is

Note I added a ? after the first " at the beginning of the pattern as well.

Warning

Doing this comes with its own problems, for example: [^>] will match any character that isn't a > so could potentially match to the end of the string and therefore you could potentially lose all of the rest of the string from your output (depending on your replacements etc.)...

Is there any reason why a > wouldn't be present?