>
>
> Collin's suggestion may sound daunting, but it's really quite easy:
Thanks guys. I took your advice. Here is my solution:
http://www.djangosnippets.org/snippets/362/
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> About the only way to do that is just grab (.*) from the url, and
> parse it in your view, looking up slugs as needed
Collin's suggestion may sound daunting, but it's really quite easy:
your urls.py can have something like
r"^(?P(?:[-\w]+/)*[-\w]+)/?$"
and then in your view:
def
About the only way to do that is just grab (.*) from the url, and
parse it in your view, looking up slugs as needed
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I'm building a site where pages can have parent pages in the form:
* Grandparent 1
* Grandparent 2
** Parent 1
*** Child 1
*** Child 2
** Parent 2
I'd like the url to contain each parent. For example, 'child 2' would be at
/grandparent-2/parent-1/child-2/
I'm not sure how to implement this in
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