Question

您好我是pdo的初学者并尝试从列代码中获取变量，然后检查变量是否与inputelement code具有相同的值。

这是错误消息：

致命错误：在第23行的C：\ xampp \ htdocs \ social \ activation1.php中调用字符串上的成员函数fetchColumn（）

<?php require_once './auth.php'; ?>
<?php
if(isset($_POST["submit"])){

    $hostname='localhost';
    $user='root';
    $password='';

    if (isset($_POST['code'])) {
        $code=$_POST['code'];
    }

    $username = ($_SESSION['user']['username']);

    try {
        $dbh = new PDO("mysql:host=$hostname;dbname=loginsystem",$user,$password);
        $dbh->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION); // <== add this line
        $sql = " SELECT code FROM user2 WHERE username = '$username'";
        $result = $sql->fetchColumn();
    }
    catch(PDOException $e)
    {
        echo $e->getMessage();
    }
    if ($result == $_POST['code']) {
        $message['success'] = 'Neuer Benutzer (' . htmlspecialchars($_POST['username']) . ') wurde angelegt, <a href="login.php">weiter zur Anmeldung</a>.';
        header('Location: http://' . $_SERVER['HTTP_HOST'] . '/social/main.php');
    } else {
        print('<pre> Please try it again. :: ');
        print('</pre>');
    }
}
?>

Answer 1

我认为您伪造查询语句$dbh->query()。用户名也写在quotes

 $sql = "SELECT code FROM user2 WHERE username = '".$username."' ";
  if ($res = $dbh->query($sql)) {// need to add this line in your code
      // then after fetchColumn
     $result = $res->fetchColumn();
   }

从mysql row pdo获取值

1 个答案: