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 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,
));
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
您可以使用template_from_string
函数渲染存储在变量中的Twig模板。
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)
));
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