如何提取图像src url省略“？”和查询字符串？

Question

I use the following regular expression to get the url of a img-Tag:

$output = preg_match_all('/<img.+src=[\'"]([^\'"]+)[\'"].*>/i', $post_single->post_content, $matches);

However, $matches gives me the following results:

http://example.com/wp-content/uploads/2013/11/dsc_842.jpg

-> This is fine.

http://example.com/wp-content/uploads/2013/11/dsc_0546.jpg?w=640

-> This is not fine.

How can I change the regular expression in order to prevent cases where ?w=640 is included in my result?

Help is much appreciated.

Thank you!

Answer 1

It is easy make it like this:

$output = preg_match_all('/<img.+src=[\'"]([^\'"?]+)[\'"?].*>/i', $post_single->post_content, $matches);

That way ([^\'"?]+)[\'"?] first matches anything beside quotes and question marks and then requires one.

For example: https://regex101.com/r/yJ1yA1/1