If you do not provide “context_object_name”, your view may look like this: <ul> {% for publisher in object_list %} <li>{{ publisher.name }}</li> {% endfor %} </ul> But if you provide like {“context_object_name”: “publisher_list”}, then you can write view like: <ul> {% for publisher in publisher_list %} <li>{{ publisher.name }}</li> {% endfor %} </ul> That means … Read more
According to the caching docs, the correct way to cache a CBV in the URLs is: from django.views.decorators.cache import cache_page url(r’^my_url/?$’, cache_page(60*60)(MyView.as_view())), Note that the answer you linked to is out of date. The old way of using the decorator has been removed (changeset).
Create a class MyUpdateView inheritted from UpdateView and override get_success_url method: class MyUpdateView(UpdateView): def get_success_url(self): pass #return the appropriate success url Also i like to pass such parameters like template_name and model inside of inheritted class view, but not in .as_view() in urls.py
I ran into a situation where I wanted something like this. Here’s what I came up with (do note that if you’re trying to use it as an update view and it can’t find the requested object, it’ll behave as a create view rather than throwing a 404): from django.views.generic.detail import SingleObjectTemplateResponseMixin from django.views.generic.edit import … Read more
No. It is just that django gives you the option to name your views in case you need to refer to them from your code, or your templates. This is useful and good practice because you avoid hardcoding urls on your code or inside your templates. Even if you change the actual url, you don’t … Read more
In my case, the problem was that I tried to use a @decorator on the class-based view as if it was a function-based view, instead of @decorating the class correctly. EDIT: From the linked page, here is a way to apply @login_required to a class-based view: from django.contrib.auth.decorators import login_required from django.utils.decorators import method_decorator @method_decorator(login_required, … Read more
maybe you could use get_success_url() method (see reference) In this case, it’d be something like: def get_success_url(self): return reverse(‘offerta_create’,args=(self.object.id,))
I can’t test it, but I bet you need return context at the end of get_context_data 🙂
I suspect class UserForm should be model form. You may want to change fields, but it should be derived from `ModelForm. So change form definition to class UserForm(forms.ModelForm): class Meta: model = User fields = […] # list of fields you want from model #or define fields that you want. ….
Have you tried passing in success_url? e.g. CreateView.as_view(model=myModel, success_url=”/success/”) or if you want to redirect to a named view: CreateView.as_view(model=myModel, success_url=reverse(‘success-url’))