尝试向数据库插入值时出错

f1tvaqid  于 2021-06-20  发布在  Mysql
关注(0)|答案(0)|浏览(177)

我正在练习php,现在,我正在学习使用mysqli将值插入数据库,但我遇到了以下错误: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 1 我不知道我做错了什么!。这是我的密码:

header('Content-Type: text/html; charset=utf-8');
ini_set("display_errors", 1);

function getUserIP()
{
    // Get real visitor IP behind CloudFlare network
    if (isset($_SERVER["HTTP_CF_CONNECTING_IP"])) {
              $_SERVER['REMOTE_ADDR'] = $_SERVER["HTTP_CF_CONNECTING_IP"];
              $_SERVER['HTTP_CLIENT_IP'] = $_SERVER["HTTP_CF_CONNECTING_IP"];
    }
    $client  = @$_SERVER['HTTP_CLIENT_IP'];
    $forward = @$_SERVER['HTTP_X_FORWARDED_FOR'];
    $remote  = $_SERVER['REMOTE_ADDR'];

    if(filter_var($client, FILTER_VALIDATE_IP))
    {
        $ip = $client;
    }
    elseif(filter_var($forward, FILTER_VALIDATE_IP))
    {
        $ip = $forward;
    }
    else
    {
        $ip = $remote;
    }

    return $ip;
}

if(isset($_GET['lat']) && !empty($_GET['lat']))
{
    $lat = $_GET['lat'];
} else {
    die();
}

if(isset($_GET['lon']) && !empty($_GET['lon']))
{
    $lon = $_GET['lon'];
} else {
    die();
}

$ip = getUserIP();
require('database.php');

$query = mysqli_query($mysqli, "SELECT * FROM `tracking` WHERE ip='".$ip."'");
if (!$query)
{
    die('Error: ' . mysqli_error($mysqli));
}
if(mysqli_num_rows($query) > 0){
    mysqli_query($mysqli,"UPDATE `tracking` SET lat='".$lat."', lon='".$lon."' WHERE ip='".$ip."'") or die(mysqli_error($mysqli)); echo "x1";
}
if(mysqli_num_rows($query) == 0){
    mysqli_query($mysqli,"INSERT INTO `tracking` (id, ip, lat, lon) VALUES (NULL, '".$ip."', '".$lat."', '".$lon."'") or die(mysqli_error($mysqli));
}
mysqli_close($mysqli);

给出错误的行是: mysqli_query($mysqli,"INSERT INTO 跟踪 (id, ip, lat, lon) VALUES (NULL, '".$ip."', '".$lat."', '".$lon."'") or die(mysqli_error($mysqli)); 我希望你能理解我想说的话。你能帮助我吗?谢谢您!

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