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