无法为用户创建安全的数据更新方式

Question

我正在为用户提供一种方法来编辑他们的数据。我的第一种方法奏效了，但后来我想起它非常不安全，我永远不应该将数据直接插入数据库；至少那是我被告知的。我尝试通过执行 VALUES (?,?,?,?,?) 来使其更安全，这样数据就不会直接进入，这在我的注册页面中似乎工作正常（如果你愿意，我可以包括在内)。

首先，这是我的原始更新数据页面，它运行良好，但没有使用 (?,?,?,?,?) 方法：

if(isset($_POST['submit'])) { 
    $userid=$_SESSION['userid'];
    $skype=$_POST['skype'];
    $email=$_POST['email'];
    $region=$_POST['region'];
    $crank=$_POST['league1'];
    $drank=$_POST['league2'];
        if(empty($skype) || empty($email) || empty($crank) || empty($drank) || empty($region))
        {
            echo "Cannot leave any field blank";
        }       
        else
        {   
            $host= "localhost";
            $dbname = "boost";
            $user = "root";
            $pwd = "";
            $port=3306; 

            try
            {
                $mysqli= new mysqli($host, $user, $pwd, $dbname,$port);
                if ($mysqli->connect_error) {
                die('Connect Error (' . $mysqli->connect_errno . ') ' . $mysqli->connect_error);
            }
            $query = "UPDATE usertable SET SkypeID = '$skype', Email = '$email', Region = '$region', CRank = '$crank', DRank = '$drank' WHERE UserID = '$userid'";
            $stmt = $mysqli->prepare($query);
            $stmt->bind_param("sssss",$skype,$email,$region,$crank,$drank); 
            $stmt->execute();
            $iLastInsertId=$mysqli->insert_id; 
            header('Location: http://localhost/Boost/account.php');
            $stmt->close(); 
            $mysqli->close();
        } catch (mysqli_sql_exception $e) { 
        throw $e; 
    } 
  }
}

这是我试图使其更安全的方法，但这似乎不起作用。具体来说，$query = "UPDATE usertable SET usertable(SkypeID,Email,Region,CRank,DRank) VALUES (?,?,?,?,?) WHERE UserID = '$userid'"; 似乎是问题所在，尽管语法对我来说看起来不错

if(isset($_POST['submit'])) { 
    $userid=$_SESSION['userid'];
    $skype=$_POST['skype'];
    $email=$_POST['email'];
    $region=$_POST['region'];
    $crank=$_POST['league1'];
    $drank=$_POST['league2'];
        if(empty($skype) || empty($email) || empty($crank) || empty($drank) || empty($region))
        {
            echo "Cannot leave any field blank";
        }       
        else
        {   
            $host= "localhost";
            $dbname = "boost";
            $user = "root";
            $pwd = "";
            $port=3306; 

            try
            {
                $mysqli= new mysqli($host, $user, $pwd, $dbname,$port);
                if ($mysqli->connect_error) {
                die('Connect Error (' . $mysqli->connect_errno . ') ' . $mysqli->connect_error);
            }
            $query = "UPDATE usertable SET usertable(SkypeID,Email,Region,CRank,DRank) VALUES (?,?,?,?,?) WHERE UserID = '$userid'";
            $stmt = $mysqli->prepare($query);
            $stmt->bind_param("sssss",$skype,$email,$region,$crank,$drank); 
            $stmt->execute();
            $iLastInsertId=$mysqli->insert_id; 
            header('Location: http://localhost/Boost/account.php');
            $stmt->close(); 
            $mysqli->close();
        } catch (mysqli_sql_exception $e) { 
        throw $e; 
    } 
  }
}

所以我不确定问题是什么。根据我使用 PHP 的经验，语法应该没问题，但我一定遗漏了一些东西。

Answer 1

其实很简单，你从

$query = "UPDATE usertable SET SkypeID = '$skype', Email = '$email', Region = '$region', CRank = '$crank', DRank = '$drank' WHERE UserID = '$userid'";

到

$query = "UPDATE usertable SET usertable(SkypeID,Email,Region,CRank,DRank) VALUES (?,?,?,?,?) WHERE UserID = '$userid'";

看来您在重写时混淆了 INSERT 语句与 UPDATE 语句，因此您只需将旧语句与新样式一起使用...

$query = "UPDATE usertable SET SkypeID = ?, Email = ?, Region = ?, CRank = ?, DRank = ? WHERE UserID = $userid";