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Redirecting to external page from django class based view

I'm trying to redirect users based on the referer in the request header. Basically, if the referer is say https://www.google.com, I would like to send them to a page, not on my website. Otherwise, continue processing as usual.
Here is what I have so far
class ArticleAccess(TemplateView, SomeMixin):
http_method_names = ['get']
template_name = 'template.html'
def dispatch(self, request, *args, **kwargs):
return super(ArticleAccess, self).dispatch(request, *args, **kwargs)
def get_context_data(self, **kwargs):
context = super(ArticleAccess, self).get_context_data(**kwargs)
item = get_object_or_404(ClientItem.objects.using(self.get_site().name), id=kwargs['article_id'])
if self.request.META.get('HTTP_REFERER') == 'https://www.google.com/':
return redirect(item.item_url)
context['id'] = item.id
context['name'] = item.name
context['html'] = item.description
context['item_url'] = item.item_url
return context
This just stays on the same page instead of redirecting. I have also tried HttpResponseRedirect, but to no avail

alecxe is correct.. you'd have to redirect from a method that is expected to return an HttpResponse.
get_context_data is not expected to return an HttpResponse and isn't ever returned by the view. It's always used to get a data dict to populate say a template. No matter what you return from this method, it will never override the response.
Therefore wherever you write this override, it needs to be in a place that is expected to return a response, such as get, post, dispatch.
The problem now is to determine how to get your object outside of the get_context_data method.

For debugging, I recommend you to start by just using a plain redirect('some_view') without conditions on your dispatch method so you can check if redirection is hit as expected and only then, go for conditions and anything else. #Yuji-tomita-tomita is just right! :) Django-pdb and ipdb are very nice tools.

How does class based view with multiple methods work with urls in django?

I have a created a view class "class Demo", which has 3 functions
def update_time()
def get_context()
def before_response()
urls.py : url(r'^demo/$', Demo.as_view(),name='demo_class'),
When i'll enter url /demo/ how will it determine which function to call from "class Demo" ?

Because Django’s URL resolver expects to send the request and associated arguments to a callable function, not a class, class-based views have an as_view() class method which serves as the callable entry point to your class. The as_view entry point creates an instance of your class and calls its dispatch() method. dispatch looks at the request to determine whether it is a GET, POST, etc, and relays the request to a matching method if one is defined, or raises HttpResponseNotAllowed if not.
just read the docs

Basically class based views are recommended when you need to handle both get and post requests at one point. For example in get method of class Register, you can render the registration form and in its post method, you can handle the form submission. If its a get request it will automatically invoke the get() method in the class, same for post request. Also you can write any common code in the dispatch() method which will be invoked for every request.eg:
class Register(View):
def dispatch(self, *args, **kwargs):
'''
common code here
'''
return super(Register, self).dispatch(*args, **kwargs)
def get(self, request):
registration_form = RegistrationForm()
return render(request, 'new.html', { 'form': registration_form })
def post(self, request):
registration_form = RegistrationForm(request.POST or None)
if registration_form.is_valid():
#save form
return HttpResponseRedirect(reverse('success-show'))
return render(request,new.html', { 'form': registration_form })

For references you can check this website.
You need to subclass a class based views, and depending on that it will have one or another method.
For example TemplateView renders a template you pass in the template_name attribute.
All class based views care about is that the attributes needed to work properly are setted. That is done via different methods. You can check the django's documentation for specifics.
For example, if you want to render a form in your template view, you will need to pass the form in the context, so you can override get_context_data() like:
def get_context_data(self, **kwargs):
context = super(DemoClass, self).get_context_data(**kwargs)
context['form'] = MyForm()
return context
There are different methods to handle different things, like querysets, objects, etc.
They are not complicated, but they are specific. If a class based view does not fit what you need, it may be easier to implement the functionality from a more general view (View, TemplateView) than forcing a more specific one to do things it is not intended for.

slightly change the url
add numbers one to three in url and put the condition in your view.
Ex.
url(r'^abc/(?P<newid>.*)$', 'class_demo'),
so your url will be like abc/1 or abc/2 or abc/3
view
def class_demo(requests, newid):
if newid==1:
update_time()
elif newid==2:
get_context()

django-cbv serve password protected static content

For every user in my django application, I generate a static .ical file which I would like to serve. Is there a CBV for that and which method would I overwrite?
As all of my views are class based I' d rather not use a function-based view with a HttpResponse and a #auth_required decorator according to this:
Django: Serving a Download in a Generic View

Just inherit from the View model and override the view method.
class ICalDownload(View):
def get(self, *args, **kwargs):
# return your response just like you would in a function view.
If you want to protect the view, I like to use django-braces. Otherwise, you need to use method_decorator on the dispatch method:
#method_decorator(auth_required)
def dispatch(self, *args, **kwargs):
return super(ICalDownload, self).dispatch(*args, **kwargs)
At this point, a function based view might be a little simpler, but like you, I like to always use class based views.

Redirect using CBV's in Django

I believe this is a simple one, just can't spot the solution. I have a view that does a bit of work on the server then passes the user back to another view, typically the original calling view.
The way I'm rendering it now, the url isn't redirected, ie it's the url of the original receiving view. So in the case the user refreshes, they'll run that server code again.
class CountSomethingView(LoginRequiredMixin, View):
def get(self, request, *args, **kwargs):
# so some counting
view = MyDetailView.as_view()
return view(request, *args, **kwargs)

I strongly recommend not overriding get or post methods. Instead, override dispatch. So, to expand on Platinum Azure's answer:
class CountSomethingView(LoginRequiredMixin, RedirectView):
permanent = False
def get_redirect_url(self, **kwargs):
url = you_can_define_the_url_however_you_want(**kwargs)
return url
def dispatch(self, request, *args, **kwargs):
# do something
return super(CountSomethingView, self).dispatch(request, *args, **kwargs)

when a user does an action and I need to redirect him to the same page, first of all I use a templateView to display a simple "thanks" (for example) then provide a link to go back to the previous page with a simple {% url %}
for example :
from django.views.generic import CreateView, TemplateView
from django.http import HttpResponseRedirect
class UserServiceCreateView(CreateView):
form_class = UserServiceForm
template_name = "services/add_service.html"
def form_valid(self, form):
[...]
return HttpResponseRedirect('/service/add/thanks/')
class UserServiceAddedTemplateView(TemplateView):
template_name = "services/thanks_service.html"
def get_context_data(self, **kw):
context = super(UserServiceAddedTemplateView, self).\
get_context_data(**kw)
context['sentance'] = 'Your service has been successfully created'
return context
in the template thanks_service.html i use {% url %} to go back to the expected page
Hope this can help

Performing a redirect in a Django Class Based View is easy.
Simply do a return redirect('your url goes here').
However, I believe this isn't what you want to do.
I see you're using get().
Normally, when speaking about HTTP, a GET request is seldom followed by a redirect.
A POST request is usually followed by a redirect because when the user goes backwards you wouldn't want to submit the same data again.
So what do you want to do?
What I think you want to do is this:
def get(self, request, *args, **kwargs):
return render_to_response('your template', data)
or even better
def get(self, request, *args, **kwargs):
return render(request, self.template_name, data)

If you're creating or updating a model, consider inheriting from CreateView or UpdateView and specifying a success_url.
If you're really doing a redirect off of an HTTP GET action, you can inherit from RedirectView and override the get method (optionally also specifying permanent = False):
class CountSomethingView(LoginRequiredMixin, RedirectView):
permanent = False
def get(self, request, *args, **kwargs):
# do something
return super(CountSomethingView, self).get(self, request, *args, **kwargs)
Note that it's really bad practice to have a get action with side-effects (unless it's just populating a cache or modifying non-essential data). In most cases, you should consider using a form-based or model-form-based view, such as CreateView or UpdateView as suggested above.

Validating a form in a get request, how?

in my way of perfectionism, I'm here to ask more questions about the not-so-well-documented class-based views.
I spend like 5 hours learning about class-based views, lurking into the code and I got a question.
Maybe what I'm trying to do is stupid, and if so, just say that.
I will put a simple example:
class SearchFormView(FormView):
template_name = 'search/search.html'
form_class = SearchForm
def get(self, request, *args, **kwargs):
form = SearchForm(self.request.GET or None)
if form.is_valid():
self.mystuff = Stuff.objects.filter(title__icontains=form.cleaned_data['query'])[:10]
return super(SearchFormView, self).get(request, *args, **kwargs)
This is a perfect valid class (it is, right?).
You have a form, and you make a GET request with a query parameter.
Works like a charm.
But lets imagine... I validate the query input to prevent some type of attack and I see that the query is malicious so I put a validation error.
With the old functions, I have a form instance (empty) and I put data in it and validation errors if needed. I always return that instance, if empty (first request) or if it filled with errors (the case of the malicious query).
The problem is with class-based views. In my get method I work with an extra instance of SearchForm so if I put validation stuff would be there and if I call get on the father it will use the instance on "form_class" that would be empty.
So, I think that there should be a way where I use the same form always, I mean: I call the request method, I pick the form_class (not create a new form), pass the data, validate and the father will return that form with the validation stuff.
Im not sure if I explained this correctly. So in short, Im creating a copy of the form in the get but I return the father get who have another copy that will be empty, so my when I display the template, there will be no errors because the form sended is empty.
Any ideas? Thanks.

Your problem is that super(SearchFormView, self).get(request, *args, **kwargs) renders its own form and own context. It's only a 3 line view function, so you should really be overriding what you need to change its behavior.
def get(self, request, *args, **kwargs):
form = SearchForm(self.request.GET or None)
if form.is_valid():
self.mystuff = Stuff.objects.filter(title__icontains=form.cleaned_data['query'])[:10]
return self.render_to_response(self.get_context_data(form=form))
Update: alternate idea if you'd like to continue using the super call
def get(self, request, *args, **kwargs):
self.form = SearchForm(self.request.GET or None)
if self.form.is_valid():
self.mystuff = Stuff.objects.filter(title__icontains=form.cleaned_data['query'])[:10]
return super(SearchFormView, self).get(request, *args, **kwargs)
def get_form(self, form_class):
"""
Returns an instance of the form to be used in this view.
"""
return getattr(self, 'form', None) or form_class(**self.get_form_kwargs())

The problem appears to be the fact that Django class based views only populate the form kwargs if the HTTP method is POST or PUT:
class FormMixin(object):
def get_form_kwargs(self):
"""
Returns the keyword arguments for instanciating the form.
"""
kwargs = {'initial': self.get_initial()}
if self.request.method in ('POST', 'PUT'):
kwargs.update({
'data': self.request.POST,
'files': self.request.FILES,
})
return kwargs
I found this a bit peculiar also, since I have on occasion used a form in a GET request (eg. a "search" form), which needed to perform some basic validation. I just override the get_form_kwargs() method on such views, to also populate the kwargs['data'] item, even when the HTTP method is GET.