Authentication 在代码点火器中按组加载模板
如何根据组ID的登录用户显示模板,以及如何根据用户组限制对控制器的访问 我的控制器:Authentication 在代码点火器中按组加载模板,authentication,acl,ion-auth,Authentication,Acl,Ion Auth,如何根据组ID的登录用户显示模板,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')
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'))
{
$this->template->client('dashboard/dashboard');
}
else
{
redirect('account/sign_in', 'refresh');
}
}
在尝试和询问之后,最终我得到了正确的结果,可以在以下代码中使用:
public function index()
{
if (!$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,并享受编写代码的时间……)