CodeIgniter：显示表单问题

Question

遵循 CodeIgniter 的优秀文档后，我有一个非常简单的新闻文章系统设置，允许任何人在访问新闻/创建时发布文章。我现在正在尝试实现一个简单的登录解决方案，它只允许登录用户发布文章，我已经按照本教程 (http://www.iluv2code.com/login-with-codeigniter-php.html) 并让它按照教程中所示的方式工作，但是当问题出现时而不是得到用户登录后登陆私人页面我希望他们登陆新闻/创建。 我试图让这个工作让我让用户登陆页面，但页面无法正确呈现，导致文章提交表单不存在。这是我认为导致问题的代码（主要是“创建”功能）：

<?php if ( ! defined('BASEPATH')) exit('No direct script access allowed');
session_start();
class News extends CI_Controller {

public function __construct()
{
    parent::__construct();
    $this->load->model('news_model');
}

public function index()
{
    $data['news'] = $this->news_model->get_news();
    $data['title'] = 'News archive';

    $this->load->view('templates/header', $data);
    $this->load->view('news/index', $data);
    $this->load->view('templates/footer');

}

public function view($slug)
{
    $data['news_item'] = $this->news_model->get_news($slug);

    if (empty($data['news_item']))
        {
            show_404();
        }

    $data['title'] = $data['news_item']['title'];

    $this->load->view('templates/header', $data);
    $this->load->view('news/view', $data);
    $this->load->view('templates/footer');
}



public function create()
{
    if($this->session->userdata('logged_in'))
        {
            $session_data = $this->session->userdata('logged_in');
            $data1['username'] = $session_data['username'];
            $this->load->view('news/create', $data1);

            $this->load->helper('form');
            $this->load->library('form_validation');

            $data['title'] = 'Create a news item';

            $this->form_validation->set_rules('title', 'Title', 'required');
            $this->form_validation->set_rules('text', 'text', 'required');

            if ($this->form_validation->run() === FALSE)
                {
                    $this->load->view('templates/header', $data);
                    $this->load->view('news/create');
                    $this->load->view('templates/footer');

                }
            else
                {
                    $this->news_model->set_news();
                    $this->load->view('templates/header', $data);
                    $this->load->view('news/success');
                    $this->load->view('templates/footer');
                }
        }

    else
        {
            //If no session, redirect to login page
            redirect('login', 'refresh');
        }
}

public function logout()
    {
        $this->session->unset_userdata('logged_in');
        session_destroy();
        redirect('news', 'refresh');
    }

}

对于那些 cmets 询问会话的人，有一个单独的控制器创建该会话（如下所示）。这运行良好，尽管在只有登录用户才能访问的“私人页面”上，它当前显示用户名，但是在初始“用户是否登录”之后，它无法呈现任何创建功能。

<?php if ( ! defined('BASEPATH')) exit('No direct script access allowed');
class VerifyLogin extends CI_Controller {
function __construct()
{
   parent::__construct();
   $this->load->model('user','',TRUE);
}

function index()
{
   //This method will have the credentials validation
   $this->load->library('form_validation');
   $this->form_validation->set_rules('username', 'Username', 'trim|required|xss_clean');
   $this->form_validation->set_rules('password', 'Password', 'trim|required|xss_clean|callback_check_database');

 if($this->form_validation->run() == FALSE)
 {
 //Field validation failed.  User redirected to login page
 $this->load->view('login_view');
 }
else
 {
 //Go to private area
 redirect('news/create', 'refresh');
 }
 }

 function check_database($password)
 {
  //Field validation succeeded.  Validate against database
  $username = $this->input->post('username');

  //query the database
  $result = $this->user->login($username, $password);

  if($result)
  {
  $sess_array = array();
  foreach($result as $row)
  {
   $sess_array = array(
     'id' => $row->id,
     'username' => $row->username
   );
   $this->session->set_userdata('logged_in', $sess_array);
  }
  return TRUE;
   }
  else
  {
 $this->form_validation->set_message('check_database', 'Invalid username or password');
 return false;
 }
 }
 }
 ?>

任何指导或帮助将不胜感激。

Answer 1

在浏览了这段代码一段时间后，我终于找到了问题，并将在此处发布解决方案，以防其他人遇到类似问题，他们可以将其用作参考。

我试图在初始检查中加载一个视图以查看用户是否已登录，删除此行允许解决方案的其余部分正常运行。

public function create()
{
if($this->session->userdata('logged_in'))
    {
        $session_data = $this->session->userdata('logged_in');
        $data1['username'] = $session_data['username'];
        $this->load->view('news/create', $data1); ----- LINE IN QUESTION

        $this->load->helper('form');
        $this->load->library('form_validation');

        $data['title'] = 'Create a news item';

        $this->form_validation->set_rules('title', 'Title', 'required');
        $this->form_validation->set_rules('text', 'text', 'required');