Question

我正在使用最新版本的symfony 3.3

我正在尝试返回json但是我收到错误

这是我的控制者：

<?php
namespace AppBundle\Controller;
use Sensio\Bundle\FrameworkExtraBundle\Configuration\Route;
use Symfony\Bundle\FrameworkBundle\Controller\Controller;
use Symfony\Component\HttpFoundation\JsonResponse;
use Symfony\Component\Serializer\SerializerInterface;

class ApiController extends Controller{
    /**
     * @Route("home", name="api_home")
     */
    public function indexAction(SerializerInterface $serializer)
    {
        $entity = $this->getDoctrine()
            ->getRepository('AppBundle:User')
            ->findAll();
        $json = $serializer->serialize($entity,'json', ['groups' => ['User']]);
        return new JsonResponse($json, 200, [], true);
    }
}

on services.yml：

services:
  _defaults:
      public: false
      autowire: false
      autoconfigure: true

config.yml

serializer:
    enabled: true
    enable_annotations: true

我收到错误：

Controller "AppBundle\Controller\ApiController::indexAction()" requires that you provide a value for the "$serializer" argument. Either the argument is nullable and no null value has been provided, no default value has been provided or because there is a non optional argument after this one.

我尝试在$ json之前死去

但同样的错误

感谢

Answer 1

这只是因为我在services.yml

上没有这个

  AppBundle\Controller\:
        resource: '../../src/AppBundle/Controller/*'
        public: true
        tags: ['controller.service_arguments']

Answer 2

action方法可以从请求转换Request属性或params（您需要从容器中检索的服务），请尝试以下代码：

/**
 * @Route("home", name="api_home")
 */
public function indexAction()
{
    $serializer = $this->get('serializer');
    $entity = $this->getDoctrine()
        ->getRepository('AppBundle:User')
        ->findAll();
    $json = $serializer->serialize($entity,'json', ['groups' => ['User']]);
    return new JsonResponse($json, 200, []);
}

希望这个帮助

symfony SerializerInterface序列化程序参数

2 个答案: