Why this url pattern is not working?

from views import login开发者_如何转开发
from django.conf.urls.defaults import *

urlpatterns = patterns('',
  ***  (r'([a-zA-Z0-9]+)/login/$', login, name='login'),

)

showing me the syntax error in line ***.

http://localhost:8000/{{ slug }}/login/ is the action url in form. I want to make it using Django { % url slug login % } tag. But it also not working.

You need to use the url() function if you're using the name keyword:

url(r'([a-zA-Z0-9]+)/login/$', login, name='login'),

And the order of parameters to the {% url %} tag has the url name first. So it should be:

{% url login slug %}

tuple literals don't have named arguments. Try url() instead.

    url(r'([a-zA-Z0-9]+)/login/$', login, name='login'),

the way u are using the {% url slug login %} is not the correct syntax you have to write the name which you wrote in your urls for that template, i.e {% url login %} This is according to your url, try this it will work.........