Django模板：将当前网址与{％url xyz％}进行比较 [英] Django Templates: Comparing current url with {% url xyz %}

问题描述

我正在尝试根据用户所在的当前页面更改导航链接的活动选择。

I am trying change the active selection of my navigation links based on the current page where the user is at.

我正在尝试执行以下操作：

I am trying to do omething like this:

<li {% if request.get_full_path == {% url profile_edit_personal %} %}class="current"{% endif %}><a href="{% url profile_edit_personal %}">Personal Details</a></li>

或者，我知道我可以定义执行以下操作：

Alternatively, I know I could define do something like this:

<li class="{% block current %}{% endblock %}"><a href="{% url profile_edit_personal %}">Personal Details</a></li>

并添加 {％block current％} current {％endblock％} 到每个相关模板，但是我更喜欢像Im在第一个示例中尝试实现的东西

and add a {% block current %}current{% endblock %} to each of the relevant templates but I would prefer something like what Im trying to achieve in the first example if possible

谢谢！

推荐答案

由于您可能只需要执行一次此操作（在您的导航模板中），对我来说保留所有内容就更有意义了

Since you'll probably only need to do this once--in your nav template--it makes more sense to me to keep everything in one place.

首先反转您的url名称，并将其存储在Timmy建议的变量中，然后只需在模板中进行比较即可：

First reverse your url names and store them in variables like Timmy suggested, then simply compare them in the template:

{% url 'about_page' as about %}
...

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

只要确保已启用请求上下文处理器，就可以访问模板中的请求。为此，可在 TEMPLATE_CONTEXT_PROCESSORS 设置变量中添加 django.core.context_processors.debug 。

Just make sure you have your request context processor enabled so you have access to the request in the template. Do this by adding django.core.context_processors.debug in your TEMPLATE_CONTEXT_PROCESSORS settings variable.

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