Hi all -
Question. I would love to use a slug in the usual way for nice urls:
#urls.py
urlpatterns = patterns('app.views',
(r'^(?P<myslug>.*)/foo/$', 'viewfunction'),
)
#views.py
def viewfunction(request, myslug):
themodel = get_object_or_404(MyModel, slug=myslug)
However, due to requirements 'themodel' will have to be pulled by a
custom id that is retrieved via an API. From there to get proper
routing I had to chang urls.py and views.py to.
#urls.py
urlpatterns = patterns('app.views',
(r'^(?P<api_id>.*)/foo/$', 'viewfunction'),
)
#views.py
def viewfunction(request, api_id):
themodel = get_object_or_404(MyModel, custom_id=api_id)
This works. But with ugly URLS:
What is the cleanest way to pull mymodel with the custom api_id, and
still have a slug/seo friendly url?
This seems simple but my mind is trying to make it complex.
Thanks,
Matt
--
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
django-users+unsubscr...@googlegroups.com.
For more options, visit this group at
http://groups.google.com/group/django-users?hl=en.
