I have a variable suppose that is: $menustr; this variable contains code html and some twig parts for example:
$menustr .= '<li><a href="{{ path("'. $actual['direccion'] .'") }}" >'. $actual['nombre'] .'</a></li>';
I need that the browser take the code html and the part of twig that in this momen is the "{{ path(~~~~~) }}"
I make a return where i send the variable called "$menustr" and after use the expresion "raw" for the html code but this dont make effective the twig code.
This is te return:
return $this->render('::menu.html.twig', array('menu' => $menustr));
and here is the template content:
{{ menu | raw }}
Twig can't render strings containing twig. There is not something like an eval function in Twig1..
What you can do is moving the path logic to the PHP stuff. The router service can generate urls, just like the path twig function does. If you are in a controller which extends the base Controller, you can simply use generateUrl:
$menuString .= '<li><a href="'.$this->generateUrl($actual['direction'].'">'. $actual['nombre'] .'</a></li>';
return $this->render('::menu.html.twig', array(
'menu' => $menuString,
));
Also, when using menu's in Symfony, I recommend to take a look at the KnpMenuBundle.
EDIT: 1. As pointed by @PepaMartinec there is a function which can do this and it is called template_from_string
You can render Twig template stored in a varible using the template_from_string function.
Check this bundle: https://github.com/LaKrue/TwigstringBundle This Bundle adds the possibility to render strings instead of files with the Symfony2 native Twig templating engine:
$vars = array('var'=>'x');
// render example string
$vars['test'] = $this->get('twigstring')->render('v {{ var }} {% if var is defined %} y {% endif %} z', $vars);
// output
v x y z
In your case i would be:
return $this->render('::menu.html.twig', array(
'menu' => $this->get('twigstring')->render($menustr, $vars)
));