Django {％url％}反向无效 [英] Django {% url %} reverse not working

问题描述

我在Django 1.4项目中有一个看法：

  def index（request）：
 print reverse 'menus_index'）
 latest_menu_list = Menu.objects.all（）。order_by（'name'）
返回render_to_response（'menus / index.html'，{'latest_menu_list'：latest_menu_list}）
  

这是按预期工作，打印出反转的URL，这是/ menus /.

在index.html模板里面（由这个视图调用）我有：

  { ％url menus_index％} 
  

哪些导致/ menus / error中的NoReverseMatch。我的应用程序的urls.py是：

  urlpatterns = patterns（'menus.views'，
 url（r'^ $'，'index'，name ='menus_index'），
 url（r'^（？P< menu_id> \d +）/ $'，'detail'，name ='menus_detail'），
）
  / pre> 
 
 
我的项目的urls.py文件中包含哪一个。
 
 
 
我做错了什么？ / p> 
 
 
更新：
 
 
 
以下是完整的index.html模板代码：
  {％extendsbase.html％} 
 {％从未来％加载url} 
 
 {％block title％ } 
菜单索引
 {％endblock％} 
 
 {％block content％} 
 {％if latest_menu_list％} 
< ul> 
 {％for menu in latest_menu_list％} 
< li>< a href ={％url menus_index％} / {{menu.id}} /> {{menu.name} }< / A>< /锂> 
 {％endfor％} 
< / ul> 
 {％else％} 
< p>没有菜单可用。< / p> 
 {％endif％} 
 {％endblock％} 
  
 
 
解决方案
答案：使用{％url'menus_index'％}。来自未来％的{％加载网址}使报价成为 https ：//docs.djangoproject.com/en/1.4/ref/templates/builtins/#url  
 
I have a view in a Django 1.4 project:
def index(request):
    print reverse('menus_index')
    latest_menu_list = Menu.objects.all().order_by('name')
    return render_to_response('menus/index.html', {'latest_menu_list': latest_menu_list})
This works as expected and prints out the reversed URL which is /menus/.

Inside of the index.html template (which is called by this view) I have:
{% url menus_index %}
Which causes a NoReverseMatch at /menus/ error.  Reverse for '' with arguments '()' and keyword arguments '{}' not found.

My application's urls.py is:
urlpatterns = patterns('menus.views',
url(r'^$','index', name='menus_index'),
url(r'^(?P<menu_id>\d+)/$','detail', name='menus_detail'),
)
Which is included in my project's urls.py file.

What am I doing wrong?

Update:

Here is the full index.html template code:
{% extends "base.html" %}
{% load url from future %}

{% block title %}
Menu Index
{% endblock %}

{% block content %}
{% if latest_menu_list %}
<ul>
    {% for menu in latest_menu_list %}
    <li><a href="{% url menus_index %}/{{ menu.id }}/">{{ menu.name }}</a></li>
    {% endfor %}
</ul>
{% else %}
   <p>No menus are available.</p>
{% endif %}
{% endblock %}

解决方案
Answer: use {% url 'menus_index' %}.  That {% load url from future %} makes the quotes a requirement per https://docs.djangoproject.com/en/1.4/ref/templates/builtins/#url

                        
这篇关于Django {％url％}反向无效的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！