Using {% url ??? %} in django templates

Instead of importing the logout_view function, you should provide a string in your file:

So not (r'^login/', login_view),

but (r'^login/', 'login.views.login_view'),

That is the standard way of doing things. Then you can access the URL in your templates using:

{% url login.views.login_view %}

Make sure (django 1.5 and beyond) that you put the url name in quotes, and if your url takes parameters they should be outside of the quotes (I spent hours figuring out this mistake!).

{% url 'namespace:view_name' arg1=value1 arg2=value2 as the_url %}
<a href="{{ the_url }}"> link_name </a>

The selected answer is out of date and no others worked for me (Django 1.6 and [apparantly] no registered namespace.)

For Django 1.5 and later (from the docs)

Warning Don’t forget to put quotes around the function path or pattern name!

With a named URL you could do:

(r'^login/', login_view, name='login'),
<a href="{% url 'login' %}">logout</a>

Just as easy if the view takes another parameter

def login(request, extra_param):
<a href="{% url 'login' 'some_string_containing_relevant_data' %}">login</a>