按组加载模板代码点火器中的Ion Auth

Question

如何根据登录用户按组id来显示模板，以及如何根据用户组限制对控制器的访问。

我的控制器：

public function index() 
{ 
    if (!$this->ion_auth->in_group('admin')) 
    { 
     $this->template->administrator('dashboard/dashboard'); 
    } 
    elseif (!$this->ion_auth->in_group('2')) 
    { 
     $this->template->admin('dashboard/dashboard'); 
    } 
    elseif (!$this->ion_auth->in_group('3')) 
    { 
     $this->template->user('dashboard/dashboard'); 
    } 
    elseif (!$this->ion_auth->in_group('members')) 
    { 
     $this->template->client('dashboard/dashboard'); 
    } 
    else 
    { 
     redirect('account/sign_in', 'refresh'); 
    } 
} 

Answer 1

努力并询问后，终于，我感到我可以如下面的代码中使用了正确的结果：

public function index() 
{ 

    if (!$this->ion_auth->logged_in()) 
    { 
     redirect('account/sign_in', 'refresh'); 
    } 
    elseif ($this->ion_auth->is_admin()) 
    { 
     $this->template->administrator('dashboard/dashboard'); 
     //View to Administrator 
    } 
    elseif($this->ion_auth->in_group('admin')) 
    { 
     $this->template->admin('dashboard/dashboard'); 
     //View to Admin 
    } 
    elseif($this->ion_auth->in_group('user')) 
    { 
     $this->template->user('dashboard/dashboard'); 
     //View to User 
    } 
    else 
    { 
     $this->template->client('dashboard/dashboard'); 
     //View to Client 
    } 
} 

希望这个答案有帮助，享受时间代码...... :)