django url - link problem
- by dana
i have an application, and in my urls.py i have something like that:
urlpatterns = patterns('',
url(r'^profile_view/(?P<id>\d+)/$',
profile_view,
name='profile_view'),)
meaning that the profile_view function has id as a parameter.
Now, i want to call that function from my template, using a link like Reply
The problem is that i don't know how to use the above url as a link, meaning how can i 'pass the id parameter to a link'?
thank you
