Different views according to url in symfony2?

1
votes

Suppose in my twig file I have following codes. Everything is working fine how to have different view if I click page number in Pagination tab. When I click page no2 it should have different output. Eg 

<p> This is page no2</p>

My url http://localhost/project/web/app_dev.php/funfact/2 When I check current url using

    {% if app.request.attributes.get('_route') == 'funfact/2' %}
</p>    this is page no 2</p>
    {%endif%}

it is not giving the above output.

How do I match current url with http://localhost/project/web/app_dev.php/funfact/2

Also in Phpstorm its giving me Alias 'Sensio\Bundle\FrameworkExtraBundle\Configuration\Route' is never used

In my twig view.html.twig

This is page 1

<div class="row">
<ul class="pagination pagination-lg">
    <li><a href="#">&laquo;</a></li>
    <li><a href="#">1</a></li>
    <li><a href="{{ path('funfact', {'page': '2'}) }}"class="active">2</a></li>
    <li><a href="{{ path('funfact', {'page': '3'}) }}"class="active">3</a></li>
    <li><a href="#">&raquo;</a></li>
</ul>
</div>

routing.yml:

funfact:
    path:     /funfact/{page}
    defaults: { _controller: HomeBundle:FunFact:funView, page: 1  }

Controller Class

class FunFactController extends Controller
{
    /**
     * @Route("/funfact/{page}"),defaults={"page" = 1})
     */
    public function funViewAction($page)
    {
      return $this->render('HomeBundle::funfact.html.twig');
    }
}
1
symfonyroutingtwig

1 Answers

1
votes

Try this code

{% if 'funfact/2' in app.request.uri  %}
    </p>this is page no 2</p>
{% endif %}