how to get url for django url include without name

Question

I have the following main site urls.py:

urlpatterns = patterns('',


   (r'^search/', include('haystack.urls')),

)

I need to call this from a link, but this type of entry has no name I can with something like:

<a href="{% url 'search???' %}">Go To Search</a>

How can I call this url entry?

petkostas · Accepted Answer · 2014-08-12 19:22:45Z

4

Django haystack urls are already named:

So you can use:

<a ref={% url "haystack_search" %}>search page</a>

answered Aug 12, 2014 at 19:17

petkostas

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Comments

user1840592 · Accepted Answer · 2014-08-12 19:14:32Z

1

you can do:

<a href='/appname/search'>Go To Search</a>

answered Aug 12, 2014 at 19:14

user1840592

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