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Codeigniter 3应用程序错误:单击浏览器的后退按钮激发signout()函数

login codeigniter php

Razvan Zamfir · 技术社区 · 7 年前

我做了一个 注册和登录 在Codeigniter 3中的应用。我一直在努力解决这个问题:登录后,用户被重定向到主页但是单击浏览器的后退按钮将触发signout()函数,而不是 重定向 从登录页面到主页。

这个 登录控制器 具有以下代码:

class Signin extends CI_Controller {

  public function __construct()
  {
    parent::__construct();
  }

  public function index()
  {
    $this->load->view('signin');
  }

  public function signin()
  {  
    $this->form_validation->set_rules('email', 'Email', 'required|trim|valid_email');
    $this->form_validation->set_rules('password', 'Password', 'required');
    $this->form_validation->set_error_delimiters('<p class="error">', '</p>');
    if ($this->form_validation->run())
    {
      $email = $this->input->post('email');
      $password = $this->input->post('password');
      $this->load->model('Usermodel');
      $current_user = $this->Usermodel->user_login($email, $password);
      // Unset the current user's data
      if ($current_user) {
       $this->session->set_userdata(
         array(
          'user_id' => $current_user->id,
          'user_email' => $current_user->email,
          'user_first_name' => $current_user->fname,
          'is_logged_in' => TRUE
          )
         );
       redirect('home');  
     } else {
        $this->session->set_flashdata("signin_failure", "Incorrect email or password");
        redirect('signin'); 
      }
    }
    else
    {
     $this->load->view('signin');
   }
  }

  public function signout(){
    // Unset the current user's data
    $this->session->unset_userdata('user_id');
    $this->session->unset_userdata('user_email');
    $this->session->unset_userdata('user_first_name');
    $this->session->unset_userdata('is_logged_in');

    // Load the signin view
    $this->load->view('signin'); 
  }
}

在导航栏(header.php视图)中,我使用 is_logged_in 显示 Sigin公司 , 登记和注销需要时的按钮:

<div id="navbar" class="collapse navbar-collapse">
    <ul class="nav navbar-nav navbar-right">
      <?php if($this->session->userdata('is_logged_in')) : ?>
      <li class="dropdown">
        <a class="dropdown-toggle" data-toggle="dropdown" href="#"><span class="glyphicon glyphicon-user"></span> Welcome, <?php echo $this->session->userdata('user_first_name'); ?>
        <span class="caret"></span></a>
        </a>
        <ul class="dropdown-menu">
          <li><a href="#"><span class="glyphicon glyphicon-phone"></span> My contacts</a></li>
          <li><a href="#"><span class="glyphicon glyphicon-phone-alt"></span> All contacts</a></li>
          <li><a href="<?php echo base_url('index.php/Signin/signout'); ?>" title="Signout"><span class="glyphicon glyphicon-log-out"></span> Signout</a></li>
        </ul>
      </li>
      <?php else: ?>
      <li><a href="#"><span class="glyphicon glyphicon-log-in"></span> Signin</a></li>
      <li><a href="#"><span class="glyphicon glyphicon-user"></span> Register</a></li>
      <?php endif;?>
    </ul>
 </div>

之后注销按钮使用一次,用户登录再次单击浏览器的后退按钮加载 http://localhost/ciauth/index.php/Signin/signout 并激发signout()函数。

我做错了什么?非常感谢。

2 回复 | 直到 7 年前

Don't Panic 7 年前

您需要将下面提到的行添加到 index & signin 方法。

if ($this->session->userdata('is_logged_in')==TRUE)
    redirect('home');

Kamarul Anuar 7 年前

问题出现在这里:

$current_user = $this->Usermodel->user_login($email, $password);

if($current_user)
{
  #is always !$current_user because $email and $password is empty. you should build condition with userdata first. if userdata not false or whatever condition comparing..and then, how you you pass output(return) in model $this->Usermodel->user_login($email, $password)??
}
else
{
  #so it going here.
  redirect('signin');
}

用新建议更新。。

public function signin() {  
   $this->form_validation->set_rules('email', 'Email', 'required|trim|valid_email');
   $this->form_validation->set_rules('password', 'Password', 'required');
   $this->form_validation->set_error_delimiters('<p class="error">', '</p>');

   if ($this->form_validation->run()) { // that's mean user come here after submit form..
      $email = $this->input->post('email');
      $password = $this->input->post('password');
      $this->load->model('Usermodel');
      $current_user = $this->Usermodel->user_login($email, $password);

      if ($current_user) { #maksure you function user_login() in Usermodel will not return FALSE if success login with $email and #password (must return something or TRUE). Then MUST return FALSE if failed to login.
         // re-set back the current user's data
         $this->session->set_userdata(
            array(
               'user_id' => $current_user->id,
               'user_email' => $current_user->email,
               'user_first_name' => $current_user->fname,
               'is_logged_in' => TRUE
            )
         );
         redirect('home'); 

      } else {

         $this->session->set_flashdata("signin_failure", "Incorrect email or password");
         redirect('signin'); 

      }
   } else { // user come here not after submit login form. possible click browsers button..

      $user_id = $this->session->userdata('is_logged_in');
      if($user_id){ // if return TRUE user already login..
         print_r('hello ' . $this->session->userdata('user_first_name') . ', youre already login.. do something below');
         #do something..
      } else {
         #you need to sign in with this form because you're not login.
         $this->load->view('signin');
      }

   }

} // end of function..