Re: urlconf for multiple page slugs
> > > 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/ --~--~-~--~~~---~--~~ 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 -~--~~~~--~~--~--~---
Re: urlconf for multiple page slugs
> 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 my_view(request, path): # iterate from grandparent -> parent -> child -> grandchild for piece in path.split('/'): do_something(piece) Not too scary and rather easy to understand (except perhaps for the regexp, but that's another matter). -tim --~--~-~--~~~---~--~~ 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 -~--~~~~--~~--~--~---
Re: urlconf for multiple page slugs
About the only way to do that is just grab (.*) from the url, and parse it in your view, looking up slugs as needed --~--~-~--~~~---~--~~ 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 -~--~~~~--~~--~--~---
urlconf for multiple page slugs
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 my urls.py without hardcoding the parent slugs. Any ideas? Here is what my get_absolute_url looks like for each page: def get_absolute_url(self): url = "/%s/" % self.slug page = self while page.parent: url = "/%s%s" % (page.parent.slug,url) page = page.parent return url --~--~-~--~~~---~--~~ 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 -~--~~~~--~~--~--~---