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

Answer 1

You could do this with a simple string replace:

$image[0] = str_replace('/wp-content/uploads/', '/', $image[0]);

Or if you want to use a regular expression:

$image[0] = preg_replace('~(http://.*?)/wp-content/uploads/(.*)~', '$1/$2', $image[0]);

Answer 2

You could do this:

$image[0] = preg_replace('!^(http://[^/]*)/[^/]*!', '$1', $image[0]);

Or you might consider just splitting the string to work on its individual components:

$parts = explode('/', $image[0]);
unset($parts[3]);
$image[0] = implode('/', $parts);

Answer 3

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