Twig 渲染函数中的 Symfony2 内部路由
我的layout.html.twig
:
{{ render(controller('AcmeDemoBundle:Page:mainmenu')) }}
Page
控制器从 Doctrine 中检索所有页面,并使用一组页面呈现 mainmenu.html.twig
.
The Page
controller retrieves all pages from the Doctrine and renders mainmenu.html.twig
with a set of pages.
我的mainmenu.html.twig
:
{% if menu_pages is defined %}
{% for page in menu_pages %}
<li class="{% if app.request.attributes.get('_internal') == '_page_show' and app.request.get('id') == page.id %}active{% endif %}"><a href="{{ path('_page_show', {id: page.id}) }}">{{ page.title|e }}</a></li>
{% endfor %}
{% endif %}
但没有显示 active
类.据我了解,问题出在内部路由中.如何解决?
But no active
class is displayed. As far as I understand the problem is in internal routing. How to fix that?
推荐答案
最好不要使用 {{ render(controller('AcmeDemoBundle:Page:mainmenu')) }}
.当您使用 services 时,它会更快速、更舒适.您可以创建一个服务,该服务将在您包含它们的任何页面上显示菜单.在服务中,您可以轻松访问当前的 _route
以添加 active
类.
Better do not use {{ render(controller('AcmeDemoBundle:Page:mainmenu')) }}
. It work more fast and comfortable when you use services instead. You can create a service which will show menu on any page where you include them. And in service you can easy get access to current _route
for add active
class.
但如果你真的需要使用 {{ render(controller('AcmeDemoBundle:Page:mainmenu')) }}
,那么你需要向他们传递当前请求,例如:
But if you really need to use {{ render(controller('AcmeDemoBundle:Page:mainmenu')) }}
, so you need pass to them a current request like:
{{ render(controller('AcmeDemoBundle:Page:mainmenu', {'request': app.request})) }}
然后在行动中将请求传递给树枝:
and then in action pass request to twig:
public function mainmenuAction($request) {
return $this->render('AcmeDemoBundle:Page:mainmenu.html.twig', array('request' => $request));
}
并在树枝中使用此请求:
and in twig use this request:
{% if menu_pages is defined %}
{% for page in menu_pages %}
<li class="{% if request.get('_route') == '_page_show' and request.get('id') == page.id %}active{% endif %}"><a href="{{ path('_page_show', {id: page.id}) }}">{{ page.title|e }}</a></li>
{% endfor %}
{% endif %}
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