• • 此问题在此处已有答案**：

What to do with mysqli problems? Errors like mysqli_fetch_array(): Argument #1 must be of type mysqli_result and such（1个答案）
九年前就关门了。
我在检查Facebook的User_id是否已经存在于我的数据库中时遇到了一些麻烦（如果不存在，它应该接受该用户为新用户，否则就加载画布应用程序）。我在我的托管服务器上运行它，没有问题，但在我的本地主机上，它给了我以下错误：
mysqli_fetch_array（）要求参数1为mysqli_result，布尔值在
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是错误的，但它已经对我的主机提供商工作，所以不可能是它!