带有recaptcha和mysql插入错误的PHP表单

说到PHP，我不在行。我试图学习并构建一个非常好的php表单，它使用recaptcha并将数据发送到mysql。我在使用mysql函数时遇到问题。我得到了错误

语法错误，意外的“；”在testing.php的第42行

当我删除它时，它会启动其他错误。如果您有任何关于此错误的帮助，我们将不胜感激

<?php
    error_reporting(E_ALL);
    ini_set('display_errors', 1);

    // keys from Google reCaptcha https://www.google.com/recaptcha/admin
    $sitekey = 'site_key';
    $secretkey = 'secret_key';

    $alert = '';
    if ($_SERVER['REQUEST_METHOD'] == 'POST')
    {
        $fullname;$username;$email;$message;$captcha;
            if(isset($_POST['fullname']))
                    $fullname=$_POST['fullname'];
            if(isset($_POST['username']))
                    $username=$_POST['username'];
            if(isset($_POST['email']))
                    $email=$_POST['email'];
            if(isset($_POST['message']))
                    $message=$_POST['message'];
            if(isset($_POST['g-recaptcha-response']))
                    $captcha=$_POST['g-recaptcha-response'];

                if(!$captcha)
                $alert = '<div class="alert alert-warning" role="alert">Please wait until the captcha protection give you a check mark.</div>';

            $response=json_decode(file_get_contents("https://www.google.com/recaptcha/api/siteverify?secret='.$secretkey.'&response=".$captcha."&remoteip=".$_SERVER['REMOTE_ADDR']));

        if($response->success==false)
        {
            /* lets set the error message for the alert... */
            if ($alert=='')
                $alert = '<div class="alert alert-danger" role="alert">Some how you have been detected has a spammer.</div>';
            }
        else
        {

            //Connecting to sql db.
            $connect = mysqli_connect("XXXX","XXXX","XXXX","XXXX");
            //Sending form data to sql db.
            mysqli_query($connect,"INSERT INTO pro_tools.Users (full_name, users_name, email, message)
            VALUES ('$_POST[post_fullname]', '$_POST[post_username]', '$_POST[post_email]', '$_POST[post_message]')";
            $alert = '<div class="alert alert-success" role="alert">Thank you for your submission!</div>';
            }
    }

?>

---

我想您这里有一个sytax错误

 $response=json_decode(file_get_contents("https://www.google.com/recaptcha/api/siteverify?secret='.$secretkey.'&response=".$captcha."&remoteip=".$_SERVER['REMOTE_ADDR']));

我会设法解决的

 $response=json_decode(file_get_contents("https://www.google.com/recaptcha/api/siteverify?secret=".$secretkey."&response=".$captcha."&remoteip=".$_SERVER['REMOTE_ADDR']));

您使用“this”而不是“this”，您将它混合了起来，您的查询有点不对劲，在结束
mysqli_查询时缺少了一个括号。但是，您很容易受到SQL注入的影响，因此应该使用准备好的语句

下面给出了一个例子

if ($stmt = mysqli_prepare($connect, "INSERT INTO pro_tools.Users (full_name, users_name, email, message) VALUES (?, ?, ?, ?)")) {
    mysqli_stmt_bind_param($stmt, "ssss", $_POST['fullname'], $_POST['username'], $_POST['email'], $_POST['message'], );
    mysqli_stmt_execute($stmt);
}

这将替换您当前的代码：

mysqli_query($connect,"INSERT INTO pro_tools.Users (full_name, users_name, email, message)
        VALUES ('$_POST[post_fullname]', '$_POST[post_username]', '$_POST[post_email]', '$_POST[post_message]')";

您提到“出现了其他错误”，我们不知道这些错误是什么-如果您需要帮助，您必须告诉我们出现了哪些错误，以及这些错误出现在哪些行上（我们不知道代码中的第42行）。
在生成插入查询时出现了一些语法错误：

用以下代码替换您的mysqli\u查询

$sql = "INSERT INTO pro_tools.Users (full_name, users_name, email, message)"
                    . "VALUES (?,?,?,?)";
            $stmt = $connect->prepare($sql);
            $stmt->bind_param("ssss", $_POST['post_fullname'], $_POST['post_username'], $_POST['post_email'], $_POST['post_message']);
            $stmt->execute();

第42至45号线为：-

mysqli_query($connect, "INSERT INTO pro_tools.Users (full_name, users_name, email, message)
VALUES (" . $_POST['post_fullname'] . "," . $_POST['post_username'] . "," . $_POST['post_email'] . "," . $_POST['post_message'] . ")");
$alert .= "<div class='alert alert-success' role='alert'>Thank you for your submission!</div>";

mysqli\u查询（$connect，“插入到pro\u tools.Users中（全名、用户名、电子邮件、消息）
值（“.$\u POST['POST\u fullname'.]”、“$\u POST['POST\u username'.]”、“$\u POST['POST\u email'.]”、“$\u POST['POST\u message'.]”）；
$alert.=“感谢您的提交！";

您有语法错误并且您必须变量才能在$alert
中有正确的html，这是第42行