如何在Symfony RoutingBundle（PHPCR）中创建具有自定义路径的路由？

Question

I'm currently researching Symfony CMF and PHPCR for a project I recently started. What I'm currently trying to figure out is how to create a Route and save it into the database. As far as I understand, I must use Symfony\Cmf\Bundle\RoutingBundle\Doctrine\Phpcr\Route and persist the element into the database. This works fine, but automatically generates a route path, which is not what I want. What I need to do is generate a custom route which links to a specific controller. Here is my code:

$em = $this->get('doctrine_phpcr.odm.document_manager');
$parent = $em->find(null, '/cms/routes');

$route = new \Symfony\Cmf\Bundle\RoutingBundle\Doctrine\Phpcr\Route();
$route->setParentDocument($parent);
$route->setName('my_route_name');
$route->setDefault('_controller', 'AppBaseBundle:Frontend/Users:index');

$em->persist($route);
$em->flush();

If i execute this code, the generated route will be /cms/routes/my_route_name. From what I can see, you could use $route->setPath('/testing');, but that generates the following exception:

Can not determine the prefix. Either this is a new, unpersisted document or the listener that calls setPrefix is not set up correctly.

Does anybody have any ideas how to solve this?

Answer 1

I managed to find a way to overcome this issue. Because in my project I also have the RouteAutoBundle, I created a class which extends \Symfony\Cmf\Bundle\RoutingBundle\Doctrine\Phpcr\Route. Inside that class I added:

/**
 * @PHPCR\Document(referenceable=true)
 */
class MenuRoute extends Route
{

    protected $url;

    /**
     * Get $this->url
     *
     * @return mixed
     */
    public function getUrl() {
        return $this->url;
    }

    /**
     * Set $this->url
     *
     * @param mixed $url
     */
    public function setUrl($url) {
        $this->url = $url;
    }
}

After that I added this to cmf_routing_auto.yml:

App\MenuBundle\Document\MenuRoute:
    uri_schema: /{getUrl}
    token_providers:
        getUrl: [content_method, { method: getUrl }]

So now one would just create an instance of MenuRoute (just like when using Route) and call the method setUrl($your_url) passing the desired url.

If anybody finds a better way, I'm opened to suggestions.