preg_replace_callback - do twice

Question

Yo, i'm trying to do this script working, but it doesn't work. How do i do it twice, the preg_replace_callback with two different functions. Thanks!

 function prepend_proxy($matches) {
   $url = (substr($_GET['url'], 0, 7) == 'http://') ? $_GET['url'] : "http://{$_GET['url']}";
    $prepend = $matches[2] ? $matches[2] : $url;
    $prepend = 'proxy2.php?url='. $prepend .'/';

    return $matches[1] . $prepend . $matches[3];
}

function imgprepend_proxy($matches2) {
   $url = (substr($_GET['url'], 0, 7) == 'http://') ? $_GET['url'] : "http://{$_GET['url']}";
    $prepend2 = $matches2[2] ? $matches2[2] : $url;
    $prepend2 = $prepend2 .'/';

    return $matches2[1] . $prepend2 . $matches2[3];
}


$new_content = preg_replace_callback(
    '|(href=[\'"]?)(https?://)?([^\'"\s]+[\'"]?)|i',
    'prepend_proxy',
    '|(src=[\'"]?)(https?://)?([^\'"\s]+[\'"]?)|i',
    'imgprepend_proxy',
    $content
);

Answer 1

$new_content = preg_replace_callback(
    '|(href=[\'"]?)(https?://)?([^\'"\s]+[\'"]?)|i',
    'prepend_proxy',
    preg_replace_callback(
        '|(src=[\'"]?)(https?://)?([^\'"\s]+[\'"]?)|i',
        'imgprepend_proxy',
        $content
    )
);