Using a name for dummy url patterns in django - django

Question

I have a project with an app named exams. Right now my index page is the index page of one of these apps. So the url patterns is something like this:
from django.conf.urls import patterns, include, url
urlpatterns = patterns('',
url(r'^', include('exams.urls', namespace="exams")),
)
However it is possible that in future I want to create a new view for my index page and change the url patterns to something like this:
from django.conf.urls import patterns, include, url
urlpatterns = patterns('',
url(r'^exams/', include('exams.urls', namespace="exams")),
url(r'^/$', 'mysite.views.index', name="index"),
)
I want to create some links in my template to index whether it loads the exams app or another view. So I tried to use the name argument for the url method(like above) but it seems it's not working. When I use {% url 'index' %} in my template it returns the following error:
Reverse for 'index' with arguments '()' and keyword arguments '{}' not found. 0 pattern(s) tried: []
UPD: I forgot to mention that I tried using the following code:
from django.conf.urls import patterns, include, url
urlpatterns = patterns('',
url(r'^', include('exams.urls', namespace="exams"), name="index"),
)
And I got the error I wrote above.
UPD2: To clarify the problem more:
Right now when I go to http://mydomain/ I'll see my app index. But maybe in the future I like to change my index page and use another view for creating it. So when I go to http://mydomain/ I'll see the page rendered by that view and by going to http://mydomain/exams I'll see my apps index pages. In that case my urls.py will be sth like the second code I wrote above in which I can easily link to index page using {% url 'index' %} tag. However for now I'm using the code I wrote in my first update. I want my templates to render {% url 'index' %} as a link to my index page(instead of using {% url 'exams:index' %} so when I later change the urls.py code I won't have to change all of my templates.

Answer 1

If I understand well, you want an index view that could be easily changed, while not changing the links pointing to it in your templates. You can use your index view as a wrapper for other views ie:
def a_view_that_will_show_all_the_exams(request):
# this is the page displayed for now as default, but this could change
# some logic here
def another_view_that_could_become_the_home_page(request):
# some logic here
def index(request):
return a_view_that_will_show_all_the_exams(request)
# or another_view_that_could_become_the_home_page
# or return exams.views.whatever_view
This way you can easily change the result of the index view while keeping your links static.

Answer 2

If I understood you correctly, right now you don't have a url with the name index, in his case it is normal that when the template renders, it tries to find a url named index, and when it doesn't find it, it fails. What you might need is create the url and redirect to another view for now, like explained here:
Redirect to named url pattern directly from urls.py in django?
UPDATE:
I'm still not completely sure of what you want (sorry about that), but I suspect what you want to do is something similar to this:
from django.conf.urls import patterns, include, url
from django.core.urlresolvers import reverse_lazy
from django.views.generic import RedirectView
urlpatterns = patterns('',
url(r'^exams/', include('exams.urls', namespace="exams")),
url(r'^/$', RedirectView.as_view(url=reverse_lazy('exams:index'), name="index"),
)
The solution that Raphael is giving you is similar, but the user will not be redirected (with both urls the user will see the same content). Maybe that is what you are looking for?