What's the recommended way of getting a generic view class from the url name?
url(r'^$', HomeView.as_view(), name='home')
So with 'home' I want to get the class HomeView.
3 Answers
Django 1.9 introduced 2 attributes to the functions that as_view() returns. view_initkwargs and view_class.
Combining this with Pickles' answer:
from django.urls import reverse, resolve
url = reverse('home')
view = resolve(url).func.view_class
1 Comment
fmoor
@IceFire, I added imports to the example.
The get_class I got from the following question: Does python have an equivalent to Java Class.forName()?
url = reverse('home')
resolver_match = resolve(url)
func = resolver_match.func
module = func.__module__
view_name = func.__name__
clss = get_class( '{0}.{1}'.format( module, view_name ) )
This is what I came up with myself I am very open to other answers.
1 Comment
challet
I haven't found something better, but it's not completly satisfying because it is not exactly fetching a view from its name. The name is not enough to perform a call to
reverse when the said url needs extra parameters (id for instance).Inspired by Pickels' answer:
import importlib
from django.core.urlresolvers import reverse, resolve
view_name = 'my_view_name'
view_func = resolve(reverse(view_name)).func
module = importlib.import_module(view_func.__module__)
view = getattr(module, view_func.__name__)
.as_view()for you to make both the resolving and the notation simpler.)