Django {% url %} when urls with parameters like: url(r'^foo//$', include(some.urls))

Question

i don\'t find any solution, how to get the url in template with the following configuration (using Django1.3):

urls.py

urlpatterns = pattern         


        

Answer 1

Yes, you can get your url like this:-

{% url 'bar-url' 1 2 %}

But note that your url configuration should actually be like this:-

urls.py

urlpatterns = patterns('',
    url(r'^/foo/(?P\d+)/, include('bar.urls')),
)

bar.urls.py

urlpatterns = patterns('',
    (r'^/bar/$, 'bar.views.index'),
    url(r'^/bar/(?P\d+)/$, 'bar.views.detail', name='bar-url'),
)

There is no foo-url unless you specifically map:-

urls.py

urlpatterns = patterns('',
    url(r'^/foo/(?P\d+)/$, 'another.views.foo', name='foo'),
    url(r'^/foo/(?P\d+)/, include('bar.urls')),
)

Note that $ means the end of the regular expression.