Symfony - generate url with parameter in controller

问题

I want to generate a Url directly in my controller. I want to user a url defined in my routing.yml file that needs a parameter.

I've found that code in the Cookbook (Routage section) :

$params = $router->match('/blog/my-blog-post');
// array('slug' => 'my-blog-post', '_controller' => 'AcmeBlogBundle:Blog:show')

$uri = $router->generate('blog_show', array('slug' => 'my-blog-post'));
// /blog/my-blog-post

But I don't understand to what is refering the $router. Obviously, it doesn't work. Is there a simple way to generate a routing url with a paramter in a controller ?

回答1:

It's pretty simple :

public function myAction()
{
    $url = $this->generateUrl('blog_show', array('slug' => 'my-blog-post'));
}

Inside an action, $this->generateUrl is an alias that will use the router to get the wanted route, also you could do this that is the same :

$this->get('router')->generate('blog_show', array('slug' => 'my-blog-post'));

回答2:

If you want absolute urls, you have the third parameter.

$product_url = $this->generateUrl('product_detail', 
    array(
        'slug' => 'slug'
    ),
    UrlGeneratorInterface::ABSOLUTE_URL
);

Remember to include UrlGeneratorInterface.

use Symfony\Component\Routing\Generator\UrlGeneratorInterface;

回答3:

Get the router from the container.

$router = $this->get('router');

Then use the router to generate the Url

$uri = $router->generate('blog_show', array('slug' => 'my-blog-post'));

回答4:

make sure your controller extends Symfony\Bundle\FrameworkBundle\Controller\Controller;

you should also check app/console debug:router in terminal to see what name symfony has named the route

in my case it used a minus instead of an underscore

i.e blog-show

$uri = $this->generateUrl('blog-show', ['slug' => 'my-blog-post']);

来源：https://stackoverflow.com/questions/20050460/symfony-generate-url-with-parameter-in-controller

标签

symfony

symfony-routing