preg_replace_callback - do 두 번

Question

요,이 스크립트를 작동 시키려고하는데 작동하지 않습니다. 두 가지 기능을 가진 preg_replace_callback을 두 번 어떻게 수행합니까? 감사!

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 
    ) 
);