preg_replace() Only Specific Part Of String

Question

I always have trouble with regex, I basically have a url, for example:

http://somedomain.com/something_here/bla/bla/bla/bla.jpg

What I need is a preg_replace() to replace the something_here with an empty string, and leave everything else in tact.

I have tried the following and it replaces the wrong parts:

$image[0] = preg_replace('/http:\/\/(.*)\/(.*)\/wp-content\/uploads\/(.*)/','$2' . '',$image[0]);

This ends up leaving only the part I want to replace, rather than actually replacing it!

Answer 1

The following code is based on the description you provided:

$url = 'http://somedomain.com/something_here/bla/bla/bla/bla.jpg';
$output = preg_replace('#^(https?://[^/]+/)[^/]+/(.*)$#', '$1$2', $url);
echo $output; // http://somedomain.com/bla/bla/bla/bla.jpg

Explanation:

• ^ : match begin of line
• ( : start matching group 1
  • https?:// : match http or https protocol
  • [^/]+ : match anything except / one or more times
  • / : match /
• ) : end matching group 1
• [^/]+ : match anything except / one or more times -/ : match /
• ( : start matching group 2
  • .* : match anything zero or more times (greedy)
• ) : end matching group 2
• $ : match end of line