Question

我将一些控制器定义为服务，我需要从路由名称获取控制器的类名。

对于非服务控制器，我可以使用路由器服务获取路由集合：

$route = $this->router->getRouteCollection()->get($routeName);
//Retrieve an object like that:

Route {
  -path: "/admin/dashboard"
  -host: ""
  -schemes: []
  -methods: []
  -defaults: array:1 [
    "_controller" => "AppBundle\Controller\Admin\AdminController::dashboardAction"
  ]
  -requirements: []
  -options: array:1 []
  -compiled: null
  -condition: ""
}

我可以使用$route["defaults"]["_controller"]访问控制器类名，所以这很好。

问题在于我的控制器作为服务，_controller属性是服务的名称，而不是Controller类（如app.controller.admin.user:listAction）我有服务的名称，但我需要有类名（{ {1}}）

我提出的唯一解决方案是从Container获取服务并在服务上使用AppBundle\Controller\Admin\UserController，但它只会对检索控制器/服务的类产生巨大的性能影响。

还有其他解决方案吗？

Answer 1

根据https://github.com/FriendsOfSymfony/FOSUserBundle/issues/2751的建议，我实现了一个缓存的映射，以便将路由名称解析为控制器类和方法。

<?php
// src/Cache/RouteClassMapWarmer.php
namespace App\Cache;

use Symfony\Component\Cache\Simple\PhpFilesCache;
use Symfony\Component\DependencyInjection\ContainerInterface;
use Symfony\Component\HttpKernel\CacheWarmer\CacheWarmerInterface;
use Symfony\Component\Routing\RouterInterface;

class RouteClassMapWarmer implements CacheWarmerInterface
{
    /** @var ContainerInterface */
    protected $container;
    /** @var RouterInterface */
    protected $router;

    public function __construct(ContainerInterface $container, RouterInterface $router)
    {
        $this->container = $container;
        $this->router = $router;
    }

    public function warmUp($cacheDirectory)
    {
        $cache = new PhpFilesCache('route_class_map', 0, $cacheDirectory);
        $controllers = [];
        foreach ($this->router->getRouteCollection() as $routeName => $route) {
            $controller = $route->getDefault('_controller');
            if (false === strpos($controller, '::')) {
                list($controllerClass, $controllerMethod) = explode(':', $controller, 2);
                // service_id gets resolved here
                $controllerClass = get_class($this->container->get($controllerClass));
            }
            else {
                list($controllerClass, $controllerMethod) = explode('::', $controller, 2);
            }
            $controllers[$routeName] = ['class' => $controllerClass, 'method' => $controllerMethod];
        }
        unset($controller);
        unset($route);
        $cache->set('route_class_map', $controllers);
    }

    public function isOptional()
    {
        return false;
    }
}

在我的RouteHelper中，读取它的实现看起来像这样

    $cache = new PhpFilesCache('route_class_map', 0, $this->cacheDirectory);
    $controllers = $cache->get('route_class_map');
    if (!isset($controllers[$routeName])) {
        throw new CacheException('No entry for route ' . $routeName . ' forund in RouteClassMap cache, please warmup first.');
    }

    if (null !== $securityAnnotation = $this->annotationReader->getMethodAnnotation((new \ReflectionClass($controllers[$routeName]['class']))->getMethod($controllers[$routeName]['method']), Security::class))
    {
        return $this->securityExpressionHelper->evaluate($securityAnnotation->getExpression(), ['myParameter' => $myParameter]);
    }

这应该比获取routeCollection快得多，并且在每个请求中针对容器解析service_id：method notated _controller-properties。

Symfony2：从服务名称或Route对象获取控制器作为服务类名

1 个答案: