Django url pattern name not working

Question

I'm trying to use url names to send a custom arg to my TemplateView.

Urls.py

urlpatterns = [
url(r'^$', index.as_view(template_name='index.html', gotodiv='whatis'), name='whatis'),
url(r'^$', index.as_view(template_name='index.html', gotodiv='whybepartofit'), name='whybepartofit')
]

Views.py

class index(TemplateView):
template_name = "index.html"
gotodiv = ''

def get_context_data(self, **kwargs):
    context = super(index, self).get_context_data(**kwargs)
    context['gotodiv'] = self.gotodiv
    return context

Template

 <li>
    <a href="{% url 'whatis' %}">What is it?</a>
 </li>
 <li>
    <a href="{% url 'whybepartofit' %}">Why be part of it?</a>
 </li>

The problem is that it doesn't care about the url name, because the patterns match in both cases. So it always goes to the first pattern (in this example, the one with the "whatis" gotodiv arg. Is there a way to configure the pattern so it only care about the name?.

Thanks

As an aside, it would be better to name your class-based-view Index. If you use lowercase index it looks like a function-based-view to other Django users. — Alasdair
– Alasdair, Commented Apr 23, 2018 at 12:47

Alasdair · Accepted Answer · 2018-04-23 19:23:56Z

1

As you have found, you can only have one URL pattern for r'^$', because Django will use the first pattern that matches.

You could change the URL to:

urlpatterns = [
    url(r'^$', index.as_view(template_name='index.html'), name='index'),
]

Then add the anchors to the links in the template:

 <li>
    <a href="{% url 'index' %}#whatis">What is it?</a>
 </li>
 <li>
    <a href="{% url 'index' %}#whybepartofit">Why be part of it?</a>
 </li>

edited Apr 23, 2018 at 19:23

answered Apr 23, 2018 at 12:46

Alasdair

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1 Comment

Luis Guanes Over a year ago

Thanks! that was the first thing i try, but didn't work then. Gave it a second try after your answer and works perfectly!

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