Django class view: init

Question

I want to get <Model> value from a URL, and use it as an __init__ parameter in my class.

urls.py
url(r'^(?P<Model>\w+)/foo/$', views.foo.as_view(), name='foo_class'),

views.py
class foo(CreateView):
    def __init__(self, **kwargs): 
        text = kwargs['Model']         # This is not working
        text = kwargs.get('Model')     # Neither this
        Bar(text)
        ...

Clearly, I'm missing something, or my understanding of URL <> class view is wrong.

there is no request if you use .as_view(). 'name 'request' is not defined' — Kotey
– Kotey, Commented Oct 6, 2017 at 6:17
You need to show more details. Where do you want to use these parameters? Usually in Django views overriding __init__ is the wrong thing to do. — Daniel Roseman
– Daniel Roseman, Commented Oct 6, 2017 at 6:36
I want to use generic view CreateView to add objects to model, and pass model name by URL. ex. To add object to Foo1: xxx/Foo1/foo, add object to Bar2: xxx/Bar2/foo. I have class ready, it works with hardcoded inputs, but I can't get it going with kwargs. — Kotey
– Kotey, Commented Oct 6, 2017 at 7:21
But where is that code? You'd probably need to put it in get_form_class. — Daniel Roseman
– Daniel Roseman, Commented Oct 6, 2017 at 8:40
For the third time. You do not need to, and must not, override __init__. Put that code in get_form_class. — Daniel Roseman
– Daniel Roseman, Commented Oct 6, 2017 at 10:58

nnyby · Accepted Answer · 2024-05-15 21:52:08Z

17

You should override dispatch method for such use cases.

class Foo(CreateView):

    def dispatch(self, request, *args, **kwargs):
        # do something extra here ...
        return super().dispatch(request, *args, **kwargs)

For your specific scenario, however, you can directly access self.kwargs as generic views automatically assign them as an instance variable on the view instance.

edited May 15, 2024 at 21:52

nnyby

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answered Oct 6, 2017 at 11:36

hspandher

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