add active class to link with sf2 and twig

Question

following simple code:

<li><a href="{{ path('_list') }}">List</a></li>

is there a simple way to add an class="active" if the current page matches the _list route?

using the newest PR-Release of symfony2 and twig as template engine

Answer 1

This is done with symfony 3.4, but probably something similar can be done with SF2.

src\AppBundle\Twig\AppExtension.php

<?php

namespace AppBundle\Twig;

use Symfony\Component\HttpFoundation\RequestStack;

class AppExtension extends \Twig_Extension
{
    private $requestStack;

    public function __construct(RequestStack $requestStack)
    {
        $this->requestStack = $requestStack;
    }

    public function getFunctions()
    {
        return [
            new \Twig_SimpleFunction('activeMenu', [$this, 'activeMenu'])
        ];
    }

    /**
     * Pass route names. If one of route names matches current route, this function returns
     * 'active'
     * @param array $routesToCheck
     * @return string
     */
    public function activeMenu(array $routesToCheck)
    {
        $currentRoute = $this->requestStack->getCurrentRequest()->get('_route');

        foreach ($routesToCheck as $routeToCheck) {
            if ($routeToCheck == $currentRoute) {
                return 'active';
            }
        }

        return '';
    }
}

Add this to services.yml

services:
    #... some other services
    AppBundle\Twig\AppExtension:
        arguments: ["@request_stack"]

Usage:

<ul class="nav navbar-nav">
    <li class="{{ activeMenu(['form', 'edit_form']) }}"><a href="{{ path('form') }}">Form</a></li>
    <li class="{{ activeMenu(['list']) }}"><a href="{{ path('list') }}">List</a></li>
</ul>

Answer 2

With ternary operator:

    {% set route = app.request.attributes.get('_route') %}
    <ul class="nav navbar-nav">
        <li {{ route ==  'profile_index' ? 'class="active"' }}><a href="{{ path('profile_index') }}"><i class="icon-profile position-left"></i> My Profile</a></li>
        <li {{ route ==  'influencers_index' ? 'class="active"'}}><a href="{{ path('influencers_index') }}"><i class="icon-crown position-left"></i> Influencers</a></li>
        <li {{ route ==  'task_manager_index' ? 'class="active"'}}><a href="{{ path('task_manager_index') }}"><i class="icon-alarm-check position-left"></i> Task Manager</a></li>
    </ul>

Answer 3

Sometimes you don't want to do exact matching of a route. For those cases, you can use the "starts with" conditional logic of twig.

As an example, lets assume you are working with books. You have the following routes: book, book_show, book_new, book_edit. You want the navigation item Book to be highlighted for any of those cases. This code would accomplish that.

<a class="{% if app.request.attributes.get('_route') starts with 'book' %}active{% endif %}">Books</a>
<a class="{% if app.request.attributes.get('_route') starts with 'author' %}active{% endif %}">Authors</a>

This example works with at least Symfony 2.3.x

Answer 4

class="class_name {% if loop.index0 == 0 %}CLASSNAME{% endif %}"