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Python Django:反向查找源的URL?_Python_Django_Django Templates_Django Urls - Fatal编程技术网

Python Django:反向查找源的URL?

Python Django:反向查找源的URL?,python,django,django-templates,django-urls,Python,Django,Django Templates,Django Urls,我在为Django生成的提要执行反向URL查找时遇到问题 我在url.py中有以下设置: feeds = { 'latest': LatestEntries, } urlpatterns = patterns('', # ... # enable feeds (RSS) url(r'^feeds/(?P<url>.*)/$', 'django.contrib.syndication.views.feed', {'feed_dict':

我在为Django生成的提要执行反向URL查找时遇到问题

我在
url.py中有以下设置:

feeds = {
    'latest': LatestEntries,
}

urlpatterns = patterns('',
    # ...
    # enable feeds (RSS)
    url(r'^feeds/(?P<url>.*)/$', 'django.contrib.syndication.views.feed',
        {'feed_dict': feeds}, name='feeds_view'),
)

但是,正如你所看到的,它显然并不干燥。我缺少什么吗?

您使用的是关键字参数,因此您应该这样传递它们:) 试试这个:

<a href="{% url feeds_view url="latest" %}">RSS feeds</a>

不幸的是,在当前的提要框架中,URL反转实际上是不可能的。好消息是feed框架已经完全重构,可以与Django的URL解析机制无缝集成。这个经过重构的提要框架将随Django 1.2一起交付,该版本将在下一个版本中发布。你可以在Rob Hudson的或a上读到它

<a href="http://{{ request.META.HTTP_HOST }}/feeds/latest">RSS feeds</a>
<a href="{% url feeds_view url="latest" %}">RSS feeds</a>