Question

我正在尝试创建一个任务管理器，我需要能够获取当前登录用户的用户ID，以便在创建任务时，用户ID也会输入到数据库中。我无法让它工作它一直告诉我call to undefined function getUser()我不认为我必须定义它，我认为它是内置的。

有人可以帮我解决这个问题，非常感谢！

这是创建任务的功能

<?php

namespace Starnes\TaskBundle\Controller;

use Symfony\Component\HttpFoundation\Request;
use Symfony\Bundle\FrameworkBundle\Controller\Controller;
use Sensio\Bundle\FrameworkExtraBundle\Configuration\Method;
use Sensio\Bundle\FrameworkExtraBundle\Configuration\Route;
use Sensio\Bundle\FrameworkExtraBundle\Configuration\Template;
use Starnes\TaskBundle\Entity\Task;
use Starnes\TaskBundle\Form\TaskType;
use Starnes\UserBundle\Entity\User;

/**
 * Task controller.
 *
 * @Route("/task")
 */
class TaskController extends Controller
{

    /**
     * Lists all Task entities.
     *
     * @Route("/", name="task")
     * @Method("GET")
     * @Template()
     */
    public function indexAction()
    {
        $em = $this->getDoctrine()->getManager();

        $entities = $em->getRepository('StarnesTaskBundle:Task')->findAll();

        return array(
            'entities' => $entities,
        );
    }
    /**
     * Creates a new Task entity.
     *
     * @Route("/", name="task_create")
     * @Method("POST")
     * @Template("StarnesTaskBundle:Task:new.html.twig")
     * 
     */
    public function createAction(Request $request)
    {
        $user = new User();
        $userID = $user->getUser()->getId();
        $entity = new Task();
        $form = $this->createCreateForm($entity);
        $form->handleRequest($request);

        if ($form->isValid()) {
            $em = $this->getDoctrine()->getManager();
            $em->persist($entity);
            $entity->setUserId($userID);
            $em->flush();

            return $this->redirect($this->generateUrl('task_show', array('id' => $entity->getId())));
        }

        return array(
            'entity' => $entity,
            'form'   => $form->createView(),
        );
    }

Answer 1

简单地写

$user = $this->getUser()->getId();

您获得当前用户的ID

Answer 2

您不需要创建新的用户对象$user = new User();，您可以从安全上下文中获取当前登录的用户对象

$user = $this->container->get('security.context')->getToken()->getUser();
$user->getId();

Answer 3

这对我有用 Symfony 3

function link(scope, element, attrs, ctrl) {            
        var unbinder = scope.$watch(function () {return ctrl[scope.model] }, function(value) {
            if (!value) return;
            scope.errorMessagesList = ctrl[scope.model].$error;
            unbinder();
        })
    }

Answer 4

您可以使用此变量：{{app.user.id}}。

如果没有连接用户ID，则显示已连接的用户ID

从FOSUserBundle中的登录用户获取用户标识

4 个答案: