我在检查数据库中是否已经存在Facebook User_id时遇到了一些困难（如果不存在，它应该接受用户为新用户，否则就加载canvas应用程序）。 我在我的主机服务器上运行它，没有问题，但在我的本地主机上，它给我以下错误:

mysqli_fetch_array()要求参数1为mysqli_result，在

下面是我的代码:

<?
$fb_id = $user_profile['id'];
$locale = $user_profile['locale'];

if ($locale == "nl_NL") {
    // Checking User Data @ WT-Database
    $check1_task = "SELECT * FROM `users` WHERE `fb_id` = " . $fb_id . " LIMIT 0, 30 ";
    $check1_res = mysqli_query($con, $check1_task);
    $checken2 = mysqli_fetch_array($check1_res);
    print $checken2;
    // If the user does not exist @ WT-Database -> insert
    if (!($checken2)) {
        $add = "INSERT INTO users (fb_id, full_name, first_name, last_name, email) VALUES ('$fb_id', '$full_name', '$first_name', '$last_name', '$email')";
        mysqli_query($con, $add);
    }
    // Double-check, the user won't be able to load the app on failure inserting to the database
    if (!($checken2)) {
        echo "Excuse us " . $first_name . ". Something went terribly wrong! Please try again later!";
        exit;
    }
} else {
    include ('sorrylocale.html');
    exit;
}

我已经读到它与我的查询错误有关，但它已经工作在我的托管提供商，所以那不可能是它！