Question

我在检查我的数据库中是否已存在Facebook User_id时遇到一些问题（如果不存在，则应该接受用户作为新用户，否则只需加载画布应用程序）。我在我的托管服务器上运行它没有问题，但在我的localhost上它给了我以下错误：

mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given in

这是我的代码：

<?
$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;
}

我读过它与我的查询错误有关，但它已经在我的托管服务提供商上工作，所以不可能！

Answer 1

该查询失败并返回false。

将其放在mysqli_query()之后，看看发生了什么。

if (!$check1_res) {
    printf("Error: %s\n", mysqli_error($con));
    exit();
}

了解更多信息：

http://www.php.net/manual/en/mysqli.error.php

mysqli_fetch_array（）期望参数1为mysqli_result，布尔值为

1 个答案: