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我正在使用php,并试图调用一个代码块内的函数,但我不知道为什么它会导致错误,但它是在稍后定义的。致命错误:调用未定义的函数,PHP
PHP代码
if(isset($_GET["updateRow"])){
$_fbURL = $_GET["ajaxified"];
if(!empty($_fbURL)){
$url = $_fbURL;
$urlParts = explode("facebook.com/", $url);
$username = GetUserIDFromUsername($urlParts[1]);
function GetUserIDFromUsername($username) {
// For some reason, changing the user agent does expose the user's UID
$options = array('http' => array('user_agent' => 'some_obscure_browser'));
$context = stream_context_create($options);
$fbsite = file_get_contents('https://www.facebook.com/' . $username, false, $context);
// ID is exposed in some piece of JS code, so we'll just extract it
$fbIDPattern = '/\"entity_id\":\"(\d+)\"/';
if (!preg_match($fbIDPattern, $fbsite, $matches)) {
throw new Exception('Unofficial API is broken or user not found');
}
return $matches[1];
}
echo "My Id is : "." ".$username;
}
}
你确定你正在解析来自** URL **的确切数据吗?确定回应。 –
当我尝试使用相同的代码,只是删除isset方法,然后它返回所需的数据,你可以尝试在一个简单的PHP,提供您的Facebook网址 – Nadeem