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