CODEWITHSUNDEEP
symfony
Sergejs Rižovs
Sergejs Rižovs

Reputation: 458

generateUrl outside controller

Is there a possibility to use generateUrl() method outside of controllers?

I tried to use it in a custom repository class with $this->get('router'), but it didn't work.

update

I've found a temporary solution here:

http://www.phamviet.net/2012/12/09/symfony-2-inject-service-as-dependency-in-to-repository/

I injected the whole service container into my repository, although it's "not recommended".

But it works for now.

update2

Injecting router instead of the whole container is probably a better idea :)

Upvotes: 14

Views: 23729

Answers (4)

kcompute
kcompute

Reputation: 1

in symfony 4 and Sylius when the FormType extends an (ex.) AbstractResourceType

class PostType extends AbstractResourceType
{

    private $router;
    public function __construct(RouterInterface $router, $dataClass, $validationGroups = [])
    {
        $this->router = $router;
        parent::__construct($dataClass, $validationGroups);

    }
}

Services.yaml :

app.post.form.type:
        class: App\Form\Admin\Post\PostType
        tags:
            - { name: form.type }
        arguments: ['@router.default', '%app.model.post.class%' ]

Upvotes: 0

r1pp3rj4ck
r1pp3rj4ck

Reputation: 1487

Inject the router itself into your EntityRepsitory (like described on Development Life blog's post Symfony 2: Injecting service as dependency into doctrine repository), then you can use $this->router->generate('acme_route');

Upvotes: 2

Gmajoulet
Gmajoulet

Reputation: 700

Don't inject the container into your repository... Really, don't !

If I were you, I would create a service and injects the router in it. In this service, I would create a method, that uses the repository and adds the needed code using the router.

That's way less dirty and easy to use/understand for another developer.

Upvotes: 11

Wouter J
Wouter J

Reputation: 41934

If you take a look in the source code of Controller::generateUrl(), you see how it's done:

$this->container->get('router')->generate($route, $parameters, $referenceType);

Basically you just enter the name of the route ($route here); if exists, some parameters ($parameters) and the type of reference (one of the constants of the UrlGeneratorInterface)

Upvotes: 18

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