在树枝模板上打印包含html和twig的变量

Question

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 }}

Answer 1

Twig can't render strings containing twig. There is not something like an eval function in Twig^1..

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,
));

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Also, when using menu's in Symfony, I recommend to take a look at the KnpMenuBundle.

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EDIT: ^1. As pointed by @PepaMartinec there is a function which can do this and it is called template_from_string