【发布时间】:2014-03-21 02:40:52
【问题描述】:
我在我的项目中使用 FOS 包。我在我的管理包中创建了一个新的控制器ChangePasswordController。这是我的代码。
namespace Pondip\AdminBundle\Controller;
use Symfony\Component\HttpFoundation\Request;
use Symfony\Bundle\FrameworkBundle\Controller\Controller;
use Symfony\Component\HttpFoundation\RedirectResponse;
use FOS\UserBundle\Controller\ChangePasswordController as BaseController;
/**
Controller managing the change password *
@author Aman
/
class ChangePasswordController extends BaseController
{
/**
This action is used for change password for admin.
@author Aman
@return render view
@throws AccessDeniedException as exception
@
*/
public function changePasswordAction()
{
$user = $this->container->get('security.context')->getToken()->getUser();
$form = $this->container->get('fos_user.change_password.form');
$formHandler = $this->container->get('fos_user.change_password.form.handler');
$process = $formHandler->process($user);
if ($process) {
$this->setFlash('fos_user_success', 'Password has been changed successfully');
$url = $this->container->get('router')->generate('pondip_admin_changepassword');
return new RedirectResponse($url);
}
return $this->container->get('templating')->renderResponse(
'PondipAdminBundle:ChangePassword:changePassword.html.'.$this->container->getParameter('fos_user.template.engine'),
array('form' => $form->createView())
);
}
}
现在我想自定义当前密码的错误信息。
【问题讨论】:
标签: php mysql symfony-2.1