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 --~--~---------~--~----~------------~-------~--~----~ You received this message because you are subscribed to the Google Groups "Django users" group. To post to this group, send email to django-users@googlegroups.com To unsubscribe from this group, send email to [EMAIL PROTECTED] For more options, visit this group at http://groups.google.com/group/django-users?hl=en -~----------~----~----~----~------~----~------~--~---