symfony - 传递给 Controller ​​的参数必须是 ContainerInterface 的实例，给定 appDevDebugProjectContainer 的实例

Question

为什么我有这个错误？

Catchable Fatal Error: Argument 1 passed to Application\Sonata\ProductBundle\Controller\ProductAdminController::__construct() must be an instance of ContainerInterface, instance of appDevDebugProjectContainer given

这是我的 services.yml:

services:
    product_admin_controller:
      class: Application\Sonata\ProductBundle\Controller\ProductAdminController
      arguments: ["@service_container"]
      tags:
            - { name: doctrine.event_listener, event: postLoad, connection: default  }

我的 Controller :

class ProductAdminController extends Controller
{
    protected $container;

    public function __construct(\ContainerInterface $container)
    {
        $this->container = $container;
    }
}

Answer 1

您必须通过“调用”选项注入(inject)容器，而不是我认为的参数:

services:
    product_admin_controller:
      class: Application\Sonata\ProductBundle\Controller\ProductAdminController
      arguments: ["@another_service_you_need"]
      tags:
            - { name: doctrine.event_listener, event: postLoad, connection: default  }
      calls:
            -   [ setContainer,["@service_container"] ]

另外，不要忘记在监听器类中创建公共(public)方法“setContainer()”。