django url - link problem

Posted by dana on Stack Overflow See other posts from Stack Overflow or by dana
Published on 2010-05-28T20:57:15Z Indexed on 2010/05/28 21:02 UTC
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django

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url

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view

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link

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parameter

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

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