Question

我正在使用symfony 2.2.3与FOSRestBundle一起实施Restful网络服务。

这是我的代码，但是当我触发以下网址时，我收到了以下错误

No route found for "GET /business/john"

这是控制器代码

use FOS\RestBundle\Controller\Annotations as Rest;
use Symfony\Component\HttpKernel\Exception\NotFoundHttpException;
use Symfony\Component\HttpFoundation\Response;
use Symfony\Component\HttpFoundation\JsonResponse;

use Symfony\Bundle\FrameworkBundle\Controller\Controller;

use Symfony\Component\Security\Core\Exception\AccessDeniedException;

use FOS\RestBundle\Controller\FOSRestController;
use FOS\RestBundle\Routing\ClassResourceInterface;

class BusinessController extends Controller {


        public function getBusinessAction($para){


          return new JsonResponse(array('name' => $para));
        } 

}

这是我的网址

http://localhost/symfony/web/app_dev.php/business/john

这是我的路线

如果我从url和函数getBusinessAction中删除参数，那么它可以工作，但不能使用参数/

_business:
    resource: "@HubHomeBundle/Controller/BusinessController.php"    
    type: rest

更新

在app/confing/routing.yml

内

   hub_home:
    resource: "@HubHomeBundle/Resources/config/routing.yml"

Answer 1

只是为了确保，这是来自HubHomeBundle / Resources / config / routing.yml？

_business:
    resource: "@HubHomeBundle/Controller/BusinessController.php"    
    type: rest

如果有，请尝试编辑并将其更改为：

_business:
    resource: HubHomeBundle\Controller\BusinessController   
    type: rest

“资源”标记的值应该是控制器类的完全限定名称，而不是文件的路径

创建参数化restful webservice时没有路由发现错误

1 个答案: