CODEWITHSUNDEEP
django
Max Vallee
Max Vallee

Reputation: 468

Django: Passing a parameter into a URL pattern name

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!

Upvotes: 1

Views: 399

Answers (1)

willeM_ Van Onsem
willeM_ Van Onsem

Reputation: 477533

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'
),

Upvotes: 2

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