Hi,
Considering the following example in the documentation:
http://www.djangoproject.com/documentation/syndication_feeds/#a-complex-example
and a url pattern like this
# ...
(r'^feeds/(?P<url>.*)/$', 'django.contrib.syndication.views.feed',
{'feed_dict': feeds}, name='feed_index'),
# ...
How can I use {% url feed_index url=??? %} in a template to generate
the link to the feed? (i.e. "/rss/beats/1424/" in the example)
The problem is that, in the template, I do not know the url, instead I
do know that this is a feed for beats and I have the beat id 1424. But
I cannot synthesize the url within the template, at least not
violating the DRY.
Regards,
oMat
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