Question

Here’s my regex code:

preg_match_all('/background[-image]*:[\s]*url\(["|\']+(.*)["|\']+\)/', $css, $matches, PREG_SET_ORDER);

It looks for CSS that looks like this:

background:url('../blah.jpg');

My problem I’m having is some CSS I scrape looks like this:

background:transparent url('../blah.jpg');
background:transparent no-repeat url('../blah.jpg');

I’m no expert when it comes to regex, so I’m wondering how I can tell it to skip anything after the colon and before URL.

Answer 1

Ths should catch all the images unless I skipped anything.

preg_match_all('~\bbackground(-image)?\s*:(.*?)\(\s*(\'|")?(?<image>.*?)\3?\s*\)~i',$str,$matches);
$images = $matches['image'];
print_r($images);