Django: Passing data to view from url dispatcher without including the data in the url?

Question

I've got my mind set on dynamically creating URLs in Django, based on names stored in database objects. All of these pages should be handled by the same view, but I would like the database object to be passed to the view as a parameter when it is called. Is that possible?

Here is the code I currently have:

places = models.Place.objects.all()
for place in places: 
    name = place.name.lower()
    urlpatterns += patterns('',
        url(r'^'+name +'/$', 'misc.views.home', name='places.'+name)
    )

Is it possible to pass extra information to the view, without adding more parameters to the URL? Since the URLs are for the root directory, and I still need 404 pages to show on other values, I can't just use a string parameter. Is the solution to give up on trying to add the URLs to root, or is there another solution?

I suppose I could do a lookup on the name itself, since all URLs have to be unique anyway. Is that the only other option?

Answer 1

I think you can pass a dictionary to the view with additional attributes, like this:

url(r'^'+name +'/$', 'misc.views.home', {'place' : place}, name='places.'+name)

And you can change the view to expect this parameter.