Php 正在验证用户和未验证用户

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。即使数据库里什么都没有。对save函数的调用来自jqueryajax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;
    }
}  

---

这可能是因为回调函数使用了一个未定义的变量。 您应该将$username更改为$title\u name，或将visa更改为$title\u name

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;
  }
}  

您可以从规则中执行此操作： $this->form_validation->set_rules（'username'，'username'，'required |是唯一的[users.username]）