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Django CMS自带的标准menu.html看起来很简单,但我无法弄清楚如何突出显示当前有子级的菜单项。Django-CMS:如何让当前活动的子页面的父项目在导航菜单中突出显示?
我尝试这样做:
{% if child.children and child.selected %} active dropdown{% endif %}
但是,对于一些奇怪的原因,不能正常工作。
下面是menu.html的全码:
{% load i18n menu_tags cache %}
{% for child in children %}
<li class="{% if child.ancestor %}ancestor{% endif %}
{% if child.selected %} active{% endif %}
{% if child.children %} dropdown{% endif %}
---> {% if child.selected and child.children %} active dropdown{% endif %}">
{% if child.children %}
<a class="dropdown-toggle" data-toggle="dropdown" href="#">
{{ child.get_menu_title }} <span class="caret"></span>
</a>
<ul class="dropdown-menu">
{% show_menu from_level to_level extra_inactive extra_active template "" "" child %}
</ul>
{% else %}
<a href="{{ child.get_absolute_url }}"><span>{{ child.get_menu_title }}</span></a>
{% endif %}
</li>
{% if class and forloop.last and not forloop.parentloop %}{% endif %}
{% endfor %}