How do you use Django URL namespaces?

Question

I'm trying to get the hang of Django URL namespaces. But I can't find any examples or documentation.

Here is what I have tried.

urls.py:

from django.conf.urls.defaults import *

urlpatterns = patterns('',
    (r'^foo/', include('sub_urls', namespace='foo', app_name='foo')),
    (r'^bar/', include('sub_urls', namespace='bar', app_name='bar')),            
)

sub_urls.py:

from django.conf.urls.defaults import patterns, url
from views import view1

urlpatterns = patterns('views',
    url(r'^(?P\d+)/$', view1, name='view1')
)

views.py:

from django.shortcuts import render_to_response

def view1(request, view_id):
    return render_to_response('view1.html', locals())

In view1.html, {% url foo:view1 3 %} outputs /foo/3, and {% url bar:view1 3 %} outputs /bar/3. This holds true whether I browse to /foo/X or /bar/X.

What I want is to be able to browse to /foo/X or /bar/X, and have {% url view1 3 %} output either /foo/3 or /bar/3, respectively.

user_132504 · Answer

I think that this isn't possible in django right now. Have a look at this message board post which references Ticket 11559. I think that you're trying to do the same thing - effectively pass an implicit parameter to the URL tag.

Also, assuming that sub_urls is from the same app both times you should make sure app_name is the same in both cases. You should only need to change namespace.

user_247590 · Answer

I realize the solution below violates the DRY principal as you have to create essentially duplicate url config files for foo and bar, however I think it should work.

urls.py:

from django.conf.urls.defaults import *

urlpatterns = patterns('',
    (r'^foo/', include('sub_urls_foo')),
    (r'^bar/', include('sub_urls_bar')),            
)

sub_urls_foo.py:

from django.conf.urls.defaults import patterns, url
from views import view1

urlpatterns = patterns('views',
    url(r'^(?P\d+)/$', view1, 'view1_foo', {'namespace': 'view1_foo'})
)

sub_urls_bar.py:

from django.conf.urls.defaults import patterns, url
from views import view1

urlpatterns = patterns('views',
    url(r'^(?P\d+)/$', view1, 'view1_bar', {'namespace': 'view1_bar'})
)

views.py:

from django.shortcuts import render_to_response

def view1(request, view_id, namespace):
    return render_to_response('view1.html', locals())

And then for the template use this:

{% url namespace 3 %}

I haven't tested the idea of using a variable in the name section of the {% url %} tag, but I think it should work.

Chase_Seibert_7679 · Answer

Here is one solution I came up with.

views.py:

from django.shortcuts import render_to_response
from django.template import RequestContext

def render_response_context(view, locals):
    request = locals["request"]
    app = "bar" if request.META["PATH_INFO"].lower().startswith("/bar") else "foo"
    return render_to_response(view, locals, 
        context_instance=RequestContext(request, current_app=app))

def view1(request, view_id):    
    return render_response_context('view1.html', locals())

view1.html:

{% load extras %}
{% namespace_url view1 3 %}

extras.py:

from django import template
from django.core.urlresolvers import reverse

register = template.Library()

@register.tag
def namespace_url(parser, token):
    tag_name, view_string, arg1 = token.split_contents()
    return NamespaceUrlNode(view_string, arg1)

class NamespaceUrlNode(template.Node):
    def __init__(self, view_string, arg1):
        self.view_string = view_string
        self.arg1 = arg1
    def render(self, context):
        return reverse("%s:%s" % (context.current_app, self.view_string), args=[self.arg1])

Basically I made sure to always pass the current_app context as either "foo" or "bar", which I calculate manually by looking at the request URL. Then I use a custom tag that resolves a URL based on current_app.

It's not very generic; "foo" and "bar" are hard-coded, and the tag can only take exactly one argument. Even with those issues fixed, this seems like a hack.

How do you use Django URL namespaces?

4 Answers