php - 验证用户和不验证用户

Question

我正在尝试验证用户名的输入字段。我创建了一个回调函数来报告用户输入是否已经作为名称存在于数据库中，但每次它都报告它确实存在。即使数据库中没有任何内容。对函数 save 的调用来自 jquery ajax 请求。

public function save()
{
    $output_status = 'Error';
    $output_title = 'Not processed';
    $output_message = 'The request was unprocessed!';

    $this->form_validation->set_rules('username', 'User Name', 'required|trim|xss_clean|callback_username_check');

    if ($this->form_validation->run())
    {
        $user_saved = $this->users_model->save_user($this->input->post('username'), $this->input->post('user_directory_name'), $this->input->post('user_status'));
        if ($user_saved)
        {
            $output_status = 'Success';
            $output_title = 'User Created';
            $output_message = 'User was successfully created!';
        }
        else
        {
            $output_title = 'User Not Created';
            $output_message = 'User was not successfully created!';
        }
    }
    else
    {
        $output_title = 'Form Not Validated';
        $output_message = validation_errors();
    }
    echo json_encode(array('status' => $output_status, 'title' => $output_title, 'message' => $output_message));  
}

public function username_check($title_name)
{
    $username_exists = $this->users_model->get_user(array('username' =>$username));
    if (is_null($username_exists))
    {
        $this->form_validation->set_message('username_check', 'The username already exists!');
        return FALSE;
    }
    else
    {
        return TRUE;
    }
}  

Answer 1

这可能是因为您的回调函数正在使用未定义的变量。您应该将 $username 更改为 $title_name 或反之亦然。

public function username_check($username)
{
  $username_exists = $this->users_model->get_user(array('username' =>$username));
  if (is_null($username_exists))
  {
    $this->form_validation->set_message('username_check', 'The username already exists!');
    return FALSE;
  }
  else
  {
    return TRUE;
  }
}  

Answer 2

你可以从规则做到这一点： $this->form_validation->set_rules('username', 'Username', 'required|is_unique[users.username]');