Django：如何获取正在呈现的模板的名称

I'm implementing a bootstrap navbar as show in this example here

我正在这个例子中显示一个bootstrap导航栏

Items in a navbar are <li>'s , the "selected" item has the attribute class="active":

导航栏中的项目为

，“selected”项目的属性为class =“active”：

  <li class="active"> <a href="#"> Link1 </a> </li>
  <li>                <a href="#"> Link2 </a> </li>

In Django these items will be within a template, which gets included by any templates that are supposed to display the navbar. I'm thinking about doing it this way:

在Django中，这些项目将位于模板中，该模板将包含任何应显示导航栏的模板。我正在考虑这样做：

<li> <a href="/"        class="{% if template_name == "home.djhtml"    %}active{% endif %}"> Home    </a> </li>
<li> <a href="about/"   class="{% if template_name == "about.djhtml"   %}active{% endif %}"> About   </a> </li>
<li> <a href="contact/" class="{% if template_name == "contact.djhtml" %}active{% endif %}"> Contact </a> </li>

I would like to know if there is a built-in way to get the template_name (that is, the template being rendered, as passed to render_to_response(), in views.py)

我想知道是否有一种内置的方法来获取template_name（即，在views.py中传递给render_to_response（）的模板）

Sure, I could explicitly add a template_name variable to render_to_response(), which would solve the problem. But thinking about DRY I feel this shouldn't be needed.

当然，我可以显式地将template_name变量添加到render_to_response（），这将解决问题。但考虑到DRY，我觉得不应该这样做。

4 个解决方案

#1

I usually use a custom template tag for this use case of adding a class to the active tab, menu item, etc.

我通常使用自定义模板标记来为活动选项卡，菜单项等添加类的用例。

@register.simple_tag
def active_page(request, view_name):
    from django.core.urlresolvers import resolve, Resolver404
    if not request:
        return ""
    try:
        return "active" if resolve(request.path_info).url_name == view_name else ""
    except Resolver404:
        return ""

And here's a snippet from the top nav:

这是来自*导航的片段：

<ul class="nav">
    <li class="{% active_page request "about" %}"><a href="{% url "about" %}">About</a></li>
    ...
</ul>

#2

I use:

我用：

class="{% if 'about' in request.path %}active{% endif %}"

It's a little shorter and more robust if the URI changes, just watch out if more than one path uses about.

如果URI发生变化，它会更短，更强大，只需注意是否有多条路径使用。

#3

There is a much faster way, without creating any custom template tag!

有一种更快的方法，无需创建任何自定义模板标签！

<ul class = 'nav'>
    <li class="{% ifequal request.path 'about/'%} active {% endifequal%}">
        <a href="{% url "about" %}">About</a>
    </li>
</ul>

just be aware about request.path. It might be with slash symbol at the beginning or without at the end of the path!

请注意request.path。它可能是在开头的斜线符号或没有在路径的末尾！

#4

Appending prog.Dusans answer for Django 1.7

为Django 1.7添加prog.Dusans答案

settings.py
TEMPLATE_CONTEXT_PROCESSORS = (
    'django.core.context_processors.request',
    'django.contrib.auth.context_processors.auth'
)

views.py
from django.shortcuts import render

def index(request):
    return render(request, 'index.html')

template
{% ifequal request.path '/pathname'%}active{% endifequal%}

Best is to add it to the base template so you need only to do it once.

最好是将其添加到基本模板，因此您只需要执行一次。

Like I this version more than Kevin Stones because you have to add almost equal code to the template and don't need a template tag after all.

就像我这个版本比Kevin Stones更多，因为你必须为模板添加几乎相同的代码，毕竟不需要模板标签。

#1