Question

除了插入数据stmt之外，一切似乎都可以正常工作。

我添加了关闭连接并添加了关闭语句。

$error = $user = $pass = "";
  if (isset($_SESSION['user'])) destroySession();

  if (isset($_POST['user']))
  {
    $user = sanitizeString($_POST['user']);
    $pass = sanitizeString($_POST['pass']);

    if ($user == "" || $pass == "")
      $error = 'Not all fields were entered<br><br>';
    else
    {
    $stmt = $connection->prepare('SELECT * FROM members WHERE user=?');
    $stmt->bind_param('s', $user);
    $stmt->execute();
    $result = $stmt->get_result();

      if ($result->num_rows)
        $error = 'That username already exists<br><br>';
      else
      {
        $hashedPwd = password_hash($pass, PASSWORD_DEFAULT);

        $stmt = $connection->prepare("INSERT INTO members (user, pass) VALUES (?,?)");
        $stmt->bind_param("ss", $user, $hashedPwd);

        $stmt->execute;
        $stmt->close();

        die('<h4>Account created</h4>Please Log in.</div></body></html>');
      }
    }
  }

  $connection->close();

我希望代码可以识别用户是否存在。但是，我不能期望用新用户来更新数据库。

Answer 1

在页面顶部放置return self.command(self.display, self.colour, self.x, self.y, self.width, self.height)将显示$ stmt-> execute存在问题；

$ stmt->执行；应该是error_reporting(E_ALL);

if语句中的准备好的语句

1 个答案: