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

Well, I usually set namespace for included urls to simplify the process:

in root urlpatterns

url(r'^articles/', include('articles.urls', namespace='articles')),

in articles/urls.py

url(r'^(\d+)/$', 'read', name='read'),

url(r'^publish/$', 'publish', name='publish'),

And then in your template you can simply type:

{% url articles:read 1 %}

or

{% url articles:publish %}

More about this here.

Answer 2

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<parameter>\d+)/, include('bar.urls')),
)

bar.urls.py

urlpatterns = patterns('',
    (r'^/bar/$, 'bar.views.index'),
    url(r'^/bar/(?P<parameter2>\d+)/$, 'bar.views.detail', name='bar-url'),
)

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

urls.py

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

Note that $ means the end of the regular expression.