php 在 CodeIgniter 中调用另一个控制器中的控制器函数
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Calling a Controller function in another Controller in CodeIgniter
提问by Roman
I have a controller "user" in my codeigniter application. This controller has a function called logged_user_only():
我的 codeigniter 应用程序中有一个控制器“用户”。这个控制器有一个函数叫做logged_user_only()：
public function logged_user_only()
{
$is_logged = $this -> is_logged();
if( $is_logged === FALSE)
{
redirect('user/login_form');
}
}
As this function calls another function called is_logged(), which just checks if the session is set, if yes it returns true, else returns false.
由于此函数调用另一个名为 的函数is_logged()，该函数仅检查会话是否已设置，如果是，则返回 true，否则返回 false。
Now if i place this function in the begining of any function within same controller, it will check if the user is not logged, it will redirect to login_formotherwise continue. This works fine.
For example,
现在，如果我将此函数放在同一控制器内任何函数的开头，它将检查用户是否未登录，它将重定向到login_form否则继续。这工作正常。例如，
public function show_home()
{
$this -> logged_user_only();
$this->load->view('show_home_view');
}
Now I would like to call this logged_user_only()function in a function of another controller to check if the user is logged in or not?
现在我想logged_user_only()在另一个控制器的函数中调用这个函数来检查用户是否登录？
PS. If this can not be done, or is not recommended, where should i place this function to access in multiple controllers? Thanks.
附注。如果这不能完成，或者不推荐，我应该把这个功能放在哪里才能在多个控制器中访问？谢谢。
回答by simnom
Why not extend the controllers so the login method is within a MY controller (within the core folder of your application) and all your other controllers extend this. For example you could have:
为什么不扩展控制器，以便登录方法位于 MY 控制器内（在您的应用程序的核心文件夹内），而您的所有其他控制器都扩展它。例如你可以有：
class MY_Controller extends CI_Controller {
public function is_logged()
{
//Your code here
}
}
and your main controllers could then extend this as follows:
然后您的主控制器可以按如下方式扩展它：
class Home_Controller extends MY_Controller {
public function show_home()
{
if (!$this->is_logged()) {
return false;
}
}
}
For further information visit: Creating Core System Classes
欲了解更多信息，请访问： 创建核心系统类
New link is here: https://www.codeigniter.com/user_guide/general/core_classes.html?highlight=core%20classes
新链接在这里：https: //www.codeigniter.com/user_guide/general/core_classes.html?highlight=core%20classes
回答by fire
Calling a controller from another is not possible with CI and not recommended.
使用 CI 无法从另一个控制器调用控制器，也不推荐使用。
Either move your logged_user_onlyinto a helper or even better a core controller that you extend all of your controllers from (MY_Controller) see http://codeigniter.com/wiki/MY_Controller_-_how_to_extend_the_CI_Controller/
要么将您移动logged_user_only到帮助程序中，或者甚至更好地将您从 ( MY_Controller)扩展所有控制器的核心控制器移动到http://codeigniter.com/wiki/MY_Controller_-_how_to_extend_the_CI_Controller/
回答by upendra allu
just load user controller as library from your controller
只需从您的控制器加载用户控制器作为库
function __construct(){
parent::__construct();
$this->load->library('../controllers/user');
}
Now, call this function of user controller any where in your controller,
现在，在控制器中的任何位置调用用户控制器的这个函数，
$this->user->logged_user_only();
回答by Muhammad Nashrullah
Function login in Controller Login
控制器登录中的功能登录
$data =array('username' => $this->input->post('username'), 'password' => $this->input >post('password')) $query = $this->db->get_where('usertable',$data)
if ($query->num_rows() == 1) {
$data = array(
'username' => $this->input->post('username'),
'logged_in' => TRUE,
'role' => "user");
$this->session->set_userdata($data);
redirect('home');
}
Function Construct in Controller user
控制器用户中的函数构造
if ($this->session->userdata('logged_in') == TRUE && $this->session->userdata('role') == "user") {} else {redirect('home');}
