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Django extend base.html along with its context

I've passed a variable to base.html from my views.py file in a context. I've extended this base.html to another several templates. The variable is only visible in the base.html and not any other extended template.
It does work if I pass the same context to each templates views.py file.
As I extended the base, shouldn't it also extend the variable? Is there any other way to get this working, or am I missing something?

When you extend a template, it inherits the html code. The context needs to be always injected by the view. If you want to pass always the same context, you need to subclass the view and not the template. You can write a mixin:
class GetContextViewMixin:
def get_context_data(self, *args, **kwargs):
return ['foo': 'foo'] # Replace with the real context
Then when you need the same context, you can use inheritance:
from django.views.generic import TemplateView
# The template of this view will obtain its context from the mixin method
class ExampleView(GetContextViewMixin, TemplateView):
template_name = 'foo.html'
If you want to extend the context in subclasses, you can override get_context_data (remember to call super):
from django.views.generic import TemplateView
class ExampleView2(GetContextViewMixin, TemplateView):
template_name = 'foo2.html'
def get_context_data(self, *args, **kwargs):
context = super().get_context_data(*args, **kwargs)
context['foo2'] = 'foo2' # Extend the context
return context

How to check for a POST method in a ListView in Django views? I'm getting a 405 error

This fails for me. It gives me a "405 Method not allowed error." It is referring to the POST method yes? How does one check for a POST? Or should I do something totally differently?
class StuffList(ListView):
template_name = "list.html"
queryset = Stuff.objects.all().order_by('-whatever')
context_object_name = 'stuff'
def get(self, request, *args, **kwargs):
if request.POST:
q = request.POST.get('q')
stuff = Stuff.objects.filter(user__icontains=stuff)
return render(request, self.template_name, {'stuff': stuff, 'q': q })
In my form, I have the token placed right inside the form action, like so:
<form action="/stuff/" method="post" name="q">
{% csrf_token %}

It's because you're making a POST request but you've only defined a GET method. If you add a POST method, it should do the trick :
def post(self, request, *args, **kwargs):
stuff = request.POST.get('q')
stuff = self.get_queryset().filter(user__icontains=stuff)
return render(request, self.template_name, {'stuff': stuff, 'q': q })
However, if you're only filtering the list, using the get is a correct solution, in this case, you have to change your form tag method="post" to method="get" and make you're GET method something like this:
def get(self, request, *args, **kwargs):
stuff = self.get_queryset()
if request.GET.get('q'):
q = request.GET.get('q')
stuff = stuff.filter(user__icontains=q)
return render(request, self.template_name, {'stuff': stuff, 'q': q })
Maybe you should look at the Django forms which can do data validation, form generation and much more.
https://docs.djangoproject.com/fr/1.8/topics/forms/

The answer given by Francois (https://stackoverflow.com/a/33876878/1126755) is very helpful. I'd like to add this more generic version of the post handler:
def post(self, request, *args, **kwargs):
self.object_list = self.get_queryset()
context = self.get_context_data()
return self.render_to_response(context)
In this way other aspects of the ListView can remain unchanged.

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.

Django DeleteView without confirmation template

I am using Django DeleteView in a template and I've created a url & view.
Is it possible to skip the process of loading the _confirm_delete template and just post the delete immediately.

DeleteView responds to POST and GET requests, GET request display confirmation template, while POST deletes instance.
You can send POST request, without confirmation with form like this:
<form method="POST" action="{% url "your_delete_url_name" %}">
{% csrf_token %}<input type="submit" value="DELETE">
</form>
If you do not want to have a link instead form button, use some javascript to make invisible form, that will be submitted on link click.
It is not good practice to use GET request for updating or deleting, but if you really insist you can shortcut get method in your class view to post, ie:
def get(self, *args, **kwargs):
return self.post(*args, **kwargs)

Or you can redefine get() method in your DeleteView:
class YourDeleteView(DeleteView):
model = YourModel
success_url = '<success_url>'
def get(self, request, *args, **kwargs):
return self.post(request, *args, **kwargs)
But be careful with that, ensure that this doesn't affect other functionality.

Yes, just change the next parameter. In your return response, make sure that the dictionary that you pass in is has something like this : { 'next': '/<your_path_here>}/' }, make sure you commit the changes before adding the next parameter. You might want to change your view's get and post functions.

Or you could only allow HTTP request method delete by routing the request directly to the delete method of your class.
from django.views.generic import DeleteView
from django.http import HttpResponseForbidden
class PostDeleteView(DeleteView):
model = Post
http_method_names = ['delete']
def dispatch(self, request, *args, **kwargs):
# safety checks go here ex: is user allowed to delete?
if request.user.username != kwargs['username']:
return HttpResponseForbidden()
else:
handler = getattr(self, 'delete')
return handler(request, *args, **kwargs)
def get_success_url(self):
username = self.kwargs.get('username')
success_url = str(reverse_lazy('post:user_home', kwargs={'username': username}))
return success_url
Let's say your URL looks like this:
path('posts/delete/<int:pk>/', PostDeleteView.as_view(), name='post_delete'),
For clarity why this works, you have to analyze the post and delete methods.
def post(self, request, *args, **kwargs):
return self.delete(request, *args, **kwargs)
def delete(self, request, *args, **kwargs):
"""
Call the delete() method on the fetched object and then redirect to the
success URL.
"""
self.object = self.get_object()
success_url = self.get_success_url()
self.object.delete()
return HttpResponseRedirect(success_url)
post just calls delete and delete gets the object and success URL, deletes the object, then redirects to the success URL you provided. pk_url_kwarg = 'pk' is why I showed the <int:pk> part in the URL.

You can override the get() method to behave exactly like the delete() method:
def get(self, request, *args, **kwargs):
return self.delete(request, *args, **kwargs)
See CCBV here: https://ccbv.co.uk/projects/Django/4.1/django.views.generic.edit/DeleteView/

All you have to do is override the get_success_url method of your delete view. Then it will directly delete the object from the DB. Eg:
class YourView(DeleteView):
model = YourModel
def get_success_url(self):
return reverse('your_redirect_view')

Django: ListView with post() method?

I am trying to process two forms in a Django class based view. The site contains a form called form (based on GET) for narrowing the list results of the ListView and the second form status_form (based on POST).
Both forms are required since the ListView returns a list of items. Form lets the user restrict the choices and status_forms lets the user flag incorrect items via a modal form (therefore it needs to be in the same template).
My trouble is that ListView does not come with the method post, however FormView does. My class List inherits from both classes, but when I execute the class I get the error message:
Attribute Error: 'List' object has no attribute 'status_form'
How should I change my implementation to allow the second form been processed via the post method?
class List(PaginationMixin, ListView, FormMixin):
model = ListModel
context_object_name = 'list_objects'
template_name = 'pages/list.html'
paginate_by = 10 #how may items per page
def get(self, request, *args, **kwargs):
self.form = ListSearchForm(self.request.GET or None,)
return super(List, self).get(request, *args, **kwargs)
def post(self, request, *args, **kwargs):
self.status_form = StatusForm(self.request.POST or None)
if self.status_form.is_valid():
...
else:
return super(List, self).post(request, *args, **kwargs)
def get_queryset(self):
# define the queryset
...
# when done, pass to object_list
return object_list
def get_context_data(self, **kwargs):
context = super(List, self).get_context_data(**kwargs)
context.update(**kwargs)
context['form'] = self.form
context['status_form'] = self.status_form # Django is complaining that status_form is not existing, result since the post method is not executed
return context

# Django is complaining that status_form does not exist,
# result since the post method is not executed
context['status_form'] = self.status_form
Because you didn't define self.status_from in the first place.
You have defined it in get_context_data, and it's accessible from there.
You can access you object from get_context_data in your post method;
context = self.get_context_data(**kwargs)
status_form = context['status_form']
Also consider that you can define your status_form directly in post method itself without getting it from self or get_context_data.
Redesign you views to separate each Form processing in separate Views then tight them with each-other.
Views redesign:
In nutshell, let each view to do one job. You can create a View just for processing your status_form and name it like StatusFormProcessView then on your List view return it on its post method
class List(ListView);
def post(self, request, *args, **kwargs):
return StatusFormView.as_view()(request) # What ever you need be pass to you form processing view
This is just an example of it, need more work to be real.
For another example; On my website index page I have a search form. when user POST or GET the search form, The processing of searching doesn't exist in my IndexView, instead I handle the whole form stuff in separate view, If form should process on GET method, I'll override get() method, If form should process on POST, I'll override post() method to send search_form data to the view that is responsible for handling of processing the search_form.
Comments response
status_form = context['status_form']
shouldn't it be
context['status_form'] = status_form
after I created it ?
You want to get status_form from context, So you need to
status_form = context['status_form']
Anyway, your form data are available on self.request.POST