Django: Passing a parameter into a URL pattern name

Question

In the example below, I'm trying to use the URL pattern name (ie, 'update_profile') instead of the absolute path (/profile/str:pk/update) in one of my views. This way if I later change my url path, I don't have to go an change each time I used the absolute path.

My question: how do I pass 'pk' into a pattern name? I've tried things like:

success_url = 'update_profile' % (pk)

But that doesn't seem to work.

urls.py

path('/profile/<str:pk>/update', views.ProfileUpdateView.as_view(), name='update_profile'),

views.py

def submit_profile_form(request, pk):
    
    if request.method == "POST":
        success_url = 'update_profile' //  <------ How do I pass 'pk' in here?
        ...
        return redirect(success_url)

Thanks for your help!

willeM_ Van Onsem · Accepted Answer · 2020-10-23 15:15:55Z

2

The redirect(…) function [Django-doc] accepts positional and named parameters, so you can write:

def submit_profile_form(request, pk):
    
    if request.method == "POST":
        success_url = 'update_profile'
        # …
        return redirect(success_url, pk=pk)

If the primary key is a number, you might want to use the pk path converter instead:

path(
    '/profile/<int:pk>/update',
    views.ProfileUpdateView.as_view(),
    name='update_profile'
),

answered Oct 23, 2020 at 15:15

willeM_ Van Onsem

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1 Comment

Max Vallee Over a year ago

Exactly what I was looking for. I knew the answer was simple. Thanks a bunch mate!

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