Question

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

urls.py

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

Included url-conf:

bar.urls.py

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

bar.views.py

def detail(request, parameter, parameter2):
    obj1 = Foo.objects.get(id=parameter)
    obj2 = Bar.objects.get(id=parameter2)

Now I try to get the url in template with:

{% url bar-url parameter=1 parameter2=2 %}

I expect to get: /bar/1/foo/2/

Is it posible to use in this case the {% url %}?

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<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.