Question

是否有可供选择的操作语法在单独的文件中调度操作 - 每个操作一个文件。原因：CVS协作的可能性更大（合并冲突更少）。我找到了这个解决方案，但是我觉得从名为Controller的类扩展一个名为Action的类感觉不舒服。（因为下面的渲染方法，我需要Controller类）。将控制器作为参数传递不起作用，因为它不是可注入服务。有一种不那么难看的方式吗？例如，在services.yml中定义规则，让操作自动从＆＃34; Controller＆＃34;类？

THX

 // ./src/AppBundle/Action/LayoutsAction.php

 <?php  

 namespace AppBundle\Action;

 use Sensio\Bundle\FrameworkExtraBundle\Configuration\Route;
 use Symfony\Component\HttpFoundation\Request; 
 use Symfony\Bundle\FrameworkBundle\Controller\Controller;
 use Symfony\Component\Serializer\Serializer;

 class LayoutsAction extends Controller {
     private $serializer;

     /**
     * The action is automatically registered as a service and dependencies are autowired.
     * Typehint any service you need, it will be automatically injected.
     */
     public function __construct(Serializer $serializer)
     {
         $this->serializer = $serializer;
     }

     /**
     * @Route("/", name="homepage")
     */
     public function __invoke(Request $request)
     {
         return $this->render('layout-editor/layouts.html.twig', [
             'initialState' => $this->serializer->normalize([
                 'layouts' => $this-get('layouts.repository.layout')-findAll()
             ])
         ]);
     }
 }

我的services.yml看起来像这样：

// ./app/config/services.yml

...

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

     AppBundle\:
         resource: '../../src/AppBundle/*'
         exclude: '../../src/AppBundle/{Entity,Repository,Tests}'

     AppBundle\Action\:
         resource: '../../src/AppBundle/Action'
         public: true

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

在Symfony 3中搜索等效的sfAction（每个文件一个动作）

0 个答案: