如何在Symfony 2中没有操作的情况下将对象传递给布局视图？

Question

I would like to create in layout dynamic menu, which will be manage from administrative panel. What i know is the object can be pass to view by the action from controller.

What i want is pass object to layout ( twig engine ) without using controller and that will work independently form controller and action.

How can i do that?

Answer 1

If you want a global variable / object in Twig, you have to implement Twig_Extension and add a new global (which will be accessible to all your templates).

A simple example would be:

1) Register a service in <bundle>/Resources/config/services.yml

services:
  acme.twig.acme_extension:
    class: Acme\DemoBundle\Twig\AcmeExtension
    tags:
      - { name: twig.extension }

2) Then create the extension:

<?php

namespace Acme\DemoBundle\Twig

class AcmeExtension extends \Twig_Extension {
  public function getGlobals() {
    return [
      'menu' => new Menu(),
    ];
  }
}

3) Then in your template:

{{ menu.render }} // Whatever

Of course, you can pass arguments to the Twig extension if needed.

Anyway, if you're trying to build a menu, maybe you should have a look at KnpMenuBundle, which helps a lot (https://github.com/KnpLabs/KnpMenuBundle)

Hope it helps!