Django/drf - delete method - django

I want to delete likes from my Likes table. for which I am making an axios call from the front end with
axios({
method: "delete",
url:http://127.0.0.1:8000/api/delete/,
params: { companyid: xyz }
})
It is supposed to delete a like that has company_id = xyz in it.
The Url looks like this
path('delete/', DeleteLikesView.as_view()),
(/api/ is included in the project's urls.py. So taken care of...)
and the DeleteLikesView -
class DeleteLikesView(DestroyAPIView):
queryset = Likes.objects.all()
serializer_class = LikesSerializer
def perform_destroy(self, request):
print(self.kwargs['companyid'])
companyid = self.kwargs['companyid']
instance = Likes.objects.get(
company_id=companyid, user_id=request.user.id)
instance.delete()
I am either being stuck with errors 403 (csrf_token error. Although I tried using csrf_exempt, no luck) or 405 method not allowed(for which I refered this. the solution in this question puts me back with 403 error)
Any help is appreciated. Thank!

Set xsrfHeaderName in request as below:
// ...
xsrfHeaderName: "X-CSRFToken",
// ...
Add CSRF_COOKIE_NAME in settings.py
CSRF_COOKIE_NAME = "XSRF-TOKEN"

You can use token authentication,instead of basicauth, if you use token auth, then csrf error will not come.
You have written
instance = Likes.objects.get(company_id=companyid, user_id=request.user.id)
in above code, user_id = request.user.id will not work cause you are not logged in the session.You are using api ,u need to provide a token to tell which user is accessing the api.

You have to use decorating the class.
To decorate every instance of a class-based view, you need to decorate the class definition itself. To do this you apply the decorator to the dispatch() method of the class.
from django.views.decorators.csrf import csrf_exempt
class DeleteLikesView(DestroyAPIView):
...
#method_decorator(csrf_exempt)
def dispatch(self, *args, **kwargs):
return super().dispatch(*args, **kwargs)
def perform_destroy(self, request):
...
See more info:
https://docs.djangoproject.com/en/2.2/topics/class-based-views/intro/#decorating-the-class

Related

Django rest framwork: DELETE without pk

I user Django Rest Framwork. I want to make a api for delete an object like this
DELETE .../items/
to delete request.user's item. (Each user can create at most one item only, and only owner can delete his item.)
I use mixins.CreateModelMixin, mixins.ListModelMixin, viewsets.GenericViewSet for list view and create. I have tried
#action(methods=['delete'], detail=False, url_path='')
def leave(self, request, *args, **kwargs):
...
but url pattern will go:
.../items/leave/$
How can I config the router or path for this? Thanks
In Django rest framework decorators, if url_path be empty strig that replace by function name. So you cannot use url_path='' as a URL path.
You can use just a simple APIView with GET method, and do what you want in that. like this:
class MyDeleteAPIView(APIView):
def get(self, request, *args, **kwargs):
# for example
try:
user = request.user
instance = SomeModel.objects.get(user=user)
instance.delete()
return Response({"message":"deleted successfuly"}, status=status.HTTP_200_OK)
except:
return Response({"message":"delete fail"}, status=status.HTTP_400_BAD_REQUEST)
now you can define your desired url:
path('delete/', MyDeleteAPIView.as_view(), name='delete'),

Delete method on Django Rest Framework ModelViewSet

i have tried to delete a single ManuscriptItem instance using Postman to perform my API requests on against the view below:
class ManuscriptViewSet(viewsets.ModelViewSet):
"""Handles creating, reading and updating items."""
authentication_classes = (TokenAuthentication,)
serializer_class = serializers.ManuscriptItemSerializer
permission_classes = (permissions.PostOwnManuscript, IsAuthenticated,)
def perform_create(self, serializer):
"""Sets the user profile to the logged in user."""
serializer.save(author=self.request.user)
def get_queryset(self):
"""
This view should return a list of all the manuscripts
for the currently authenticated user.
"""
user = self.request.user
return models.ManuscriptItem.objects.filter(author=user)
def destroy(self, request, *args, **kwargs):
instance = self.get_object()
self.perform_destroy(instance)
return Response(status=status.HTTP_204_NO_CONTENT)
def perform_destroy(self, instance):
instance.delete()
The destroy and perform destroy functions are what I have attempted without success. This is what it returns when i tried:
{
"detail": "Method \"DELETE\" not allowed." }
This is how my URLs are currently registered:
router = DefaultRouter()
router.register('manuscripts', views.ManuscriptViewSet, base_name="manuscripts") # auto basename for models
router.register('manuscriptlibrary', views.ManuscriptLibraryViewSet, base_name="manuscript_library")
router.register('manuscriptsettings', views.ManuscriptSettingsViewSet)
urlpatterns = [
url(r'', include(router.urls))
]
I'm i modifying the ModelViewSet wrong do i need to use another approach because of the nature of ModelViewSet? i expected it to work on Postman when i used an Authorized user to Delete a ManuscriptItem instance. In the docs it said Destroy() method can be used.
Additional information
The URL used is:
http://localhost:8000/manuscripts-api/manuscripts/
The model instance to be deleted from:
class ManuscriptItem(models.Model):
"""Represents a single manuscript's content"""
author = models.ForeignKey('accounts_api.UserProfile', on_delete=models.CASCADE)
title = models.CharField(max_length=255)
content = models.CharField(max_length=99999999)
def __str__(self):
"""Django uses when it needs to convert the object to a string"""
return str(self.id)
The way i have tried sending delete requests on postman with json:
{
"manuscript": 7,
}
Results: Delete Method not allowed
{
"id": 7,
"author": 5,
"title": "niceone",
"content": "niceone"
}
Results: Delete Method not allowed
Additional Questions/Info:
Don't i need to specify the router register with a pk? I tried this but didnt work either:
router.register('manuscripts/{pk}/$', views.ManuscriptViewSet, base_name="manuscript_detail")
Postman says:
Allow →GET, POST, HEAD, OPTIONS
The issue here is that you send DELETE request to the wrong url. Look at the DefaultRouter docs. It generates automatically your urls within your viewset:
Look closely at the DELETE method. It is on the {prefix}/{lookup}/[.format] url pattern. This means that your corresponding router url is manuscripts/<manuscript_id>/, but you try to send DELETE request to manuscripts/ only, which is the above pattern. You see directly from the table that the allowed HTTP methods there are GET and POST only. That's why you receive MethodNotAllowed.
The solution to your problem is not to pass the manuscript_id as a JSON body of the request
{
"manuscript": 7,
}
But to pass it directly to the url:
DELETE http://localhost:8000/manuscripts-api/manuscripts/7/
And you just register your viewset like:
router.register(r'manuscripts', ManuscriptViewSet.as_view(), name='manuscripts')
As you see, DRF generates the urls automatically for you.
from rest_framework.response import Response
from rest_framework import status
def destroy(self, request, *args, **kwargs):
try:
instance = self.get_object()
self.perform_destroy(instance)
except Http404:
pass
return Response(status=status.HTTP_204_NO_CONTENT)
use this and it will work

Permission checks in DRF viewsets are not working right

I am implementing an API where I have nested structures.
Lets say it is a zoo and I can call GET /api/cage/ to get a list of cages GET /api/cage/1/ to get cage ID 1, but then I can GET /api/cage/1/animals/ to get a list of animals in that cage.
The problem I am having is with permissions. I should only be able to see animals in the cage if I can see the cage itself. I should be able to see the cage itself if has_object_permission() returns True in the relevant permission class.
For some reason, has_object_permission() gets called when I do GET /api/cage/1/, but has_permission() gets called when I call GET /api/cage/1/animals/. And with has_permission() I don't have access to the object to check the permissions. Am I missing something? How do I do this?
My cage viewset looks more or less like this
class CageViewSet(ModelViewSet):
queryset = Cage.objects.all()
serializer_class = CageSerializer
permission_classes = [GeneralZooPermissions, ]
authentication_classes = [ZooTicketCheck, ]
def get_queryset(self):
... code to only list cages you have permission to see ...
#detail_route(methods=['GET'])
def animals(self, request, pk=None):
return Request(AnimalSerializer(Animal.objects.filter(cage_id=pk), many=True).data)
My GeneralZooPermissions class looks like this (at the moment)
class GeneralZooPermissions(BasePermission):
def has_permission(self, request, view):
return True
def has_object_permission(self, request, view, obj):
return request.user.has_perm('view_cage', obj)
It seems like this is a bug in DRF. Detailed routes do not call the correct permission check. I have tried reporting this issue to DRF devs, but my report seems to have disappeared. Not sure what to do next. Ideas?
The issue I posted with DRF is back and I got a response. Seems like checking only has_permission() and not has_object_permission() is the intended behavior. This doesn't help me. At this point, something like this would have to be done:
class CustomPermission(BasePermission):
def has_permission(self, request, view):
"""we need to do all permission checking here, since has_object_permission() is not guaranteed to be called"""
if 'pk' in view.kwargs and view.kwargs['pk']:
obj = view.get_queryset()[0]
# check object permissions here
else:
# check model permissions here
def has_object_permission(self, request, view, obj):
""" nothing to do here, we already checked everything """
return True
OK, so after reading a bunch of DRF's code and posting an issue at the DRF GitHub page.
It seems that has_object_permission() only gets called if your view calls get_object() to retrieve the object to be operated on.
It makes some sense since you would need to retrieve the object to check permissions anyway and if they did it transparently it would add an extra database query.
The person who responded to my report said they need to update the docs to reflect this. So, the idea is that if you want to write a custom detail route and have it check permissions properly you need to do
class MyViewSet(ModelViewSet):
queryset = MyModel.objects.all()
....
permission_classes = (MyCustomPermissions, )
#detail_route(methods=['GET', ])
def custom(self, request, pk=None):
my_obj = self.get_object() # do this and your permissions shall be checked
return Response('whatever')
If you want to define permissions while doing another method that doesn't call the get_object() (e.g. a POST method), you can do overriding the has_permission method. Maybe this answer can help (https://stackoverflow.com/a/52783914/12737833)
Another thing you can do is use the check_object_permissions inside your POST method, that way you can call your has_object_permission method:
#action(detail=True, methods=["POST"])
def cool_post(self, request, pk=None, *args, **kwargs):
your_obj = self.get_object()
self.check_object_permissions(request, your_obj)
In my case I didn't address requests correctly so my URL was api/account/users and my mistake was that I set URL in frontend to api/account/ and thats not correct!

Adding more views to a Router or viewset (Django-Rest-Framework)

Essentially, I'm trying to find a good way to attach more views to a Router without creating a custom Router. What's a good way to accomplish this?
Here is something sort of equivalent to what I'm trying to accomplish. Variable names have been changed and the example method I want to introduce is extremely simplified for the sake of this question.
Router:
router = routers.SimpleRouter(trailing_slash=False)
router.register(r'myobjects', MyObjectViewSet, base_name='myobjects')
urlpatterns = router.urls
ViewSet
class MyObjectsViewSet(viewsets.ViewSet):
""" Provides API Methods to manage MyObjects. """
def list(self, request):
""" Returns a list of MyObjects. """
data = get_list_of_myobjects()
return Response(data)
def retrieve(self, request, pk):
""" Returns a single MyObject. """
data = fetch_my_object(pk)
return Response(data)
def destroy(self, request, pk):
""" Deletes a single MyObject. """
fetch_my_object_and_delete(pk)
return Response()
One example of another method type I need to include. (There are many of these):
def get_locations(self, request):
""" Returns a list of location objects somehow related to MyObject """
locations = calculate_something()
return Response(locations)
The end-result is that the following URL would work correctly and be implemented 'cleanly'.
GET example.com/myobjects/123/locations
The answer given by mariodev above is correct, as long as you're only looking to make GET requests.
If you want to POST to a function you're appending to a ViewSet, you need to use the action decorator:
from rest_framework.decorators import action, link
from rest_framework.response import Response
class MyObjectsViewSet(viewsets.ViewSet):
# For GET Requests
#link()
def get_locations(self, request):
""" Returns a list of location objects somehow related to MyObject """
locations = calculate_something()
return Response(locations)
# For POST Requests
#action()
def update_location(self, request, pk):
""" Updates the object identified by the pk """
location = self.get_object()
location.field = update_location_field() # your custom code
location.save()
# ...create a serializer and return with updated data...
Then you would POST to a URL formatted like:
/myobjects/123/update_location/
http://www.django-rest-framework.org/api-guide/viewsets/#marking-extra-actions-for-routing has more information if you're interested!
You can now do this with the list_route and detail_route decorators: http://www.django-rest-framework.org/api-guide/viewsets/#marking-extra-actions-for-routing
For example:
from rest_framework.decorators import list_route
from rest_framework.response import Response
...
class MyObjectsViewSet(viewsets.ViewSet):
...
#list_route()
def locations(self, request):
queryset = get_locations()
serializer = LocationSerializer(queryset, many=True)
return Response(serializer.data)
You define method like you do now, but you need to use the same url as method name and add link decorator, so for
/myobjects/123/locations/
You add method like this
#link(permission_classes=[...])
def locations(self, request, pk=None):
...
and router will pick it automatically.
From Routing to extra methods on a ViewSet:
I think you may need to route the method by hand, i.e. The Old-Fashioned Way™.
First pull the method out as a separate view:
set_password_view = UserViewSet.as_view({'post': 'set_password'})
(or such)
Then assign your URL:
url(r'^users/username_available/$', set_password_view, name-=...)
(Or such)
There's a related question on SO.
If you want to extend a viewset with a view that is or should not directly be written inside your viewset, you can write a “wrapper” action to pass the data through.
For example, with class based views:
from somewhere import YourExternalClassView
class SomeViewSet(viewsets.ReadOnlyModelViewSet):
# ...
#action(detail=True)
def your_action(self, request, pk):
return YourExternalClassView.as_view()(request, pk=pk)
How does it work?
On class based views, the as_view method returns a view function, to which we will pass the data we received from the action. The view will then hand over to process further.
For non-class based view, the views can be called/wrapped directly without .as_view(...)(...).

TemplateView in Django with custom status code

I have internal account privacy permissions in my project(e.g. only friends can see profile page of user) and I want to have custom permission denied page for this case. Is there any way to return response from TemplateView with status code equals 403?
Something like this:
class PrivacyDeniedView(TempateView):
template_name = '...'
status_code = 403
I can do this by override dispatch() but maybe Django has out of the box solution
Answer: it looks like there no generic solution. The best way is proposed by #alecxe, but encapsulated in Mixin as #FoxMaSk proposed
One option is to override get() method of your TemplateView class:
def get(self, request, *args, **kwargs):
context = self.get_context_data(**kwargs)
return self.render_to_response(context, status=403)
You can subclass TemplateResponse and set response_class in the view. For example:
from django.template.response import TemplateResponse
class TemplateResponseForbidden(TemplateResponse):
status_code = 403
class PrivacyDeniedView(TemplateView):
response_class = TemplateResponseForbidden
...
This approach is more DRY than the other suggestions because you don't need to copy and paste any code from TemplateView (e.g. the call to render_to_string()).
I tested this in Django 1.6.
While alecxe's answer works, I strongly suggest you to avoid overriding get; it's easy to forget that CBV's can have other methods like post, and if you're overriding one you should do the same for the others.
In fact, there is no need to create a separate view just to display a 403 error; Django already has django.http.HttpResponseForbidden. So instead of redirecting to your view, just do something along the lines of:
if not user.has_permission(): # or however you check the permission
return HttpResponseForbidden()
Or, if you want to render a particular template:
if not user.has_permission(): # or however you check the permission
return HttpResponseForbidden(loader.render_to_string("403.html"))
I just encountered this problem as well. My goal was to be able to specify the status code in urls.py, e.g.:
url(r'^login/error/?$', TemplateView.as_view(template_name='auth/login_error.html', status=503), name='login_error'),
So using the previous answers in this thread as idea starters, I came up with the following solution:
class TemplateView(django.views.generic.TemplateView):
status = 200
def render_to_response(self, context, **response_kwargs):
response_kwargs['status'] = self.status
return super(TemplateView, self).render_to_response(context, **response_kwargs)
This is an old question, but I came across it first when searching. I was using a different generic view (ListView) whose implementation for get is a bit more complex, and thus a bit of a mess to override. I went for this solution instead:
def get(self, request, *args, **kwargs):
response = super(ListView, self).get(request, *args, **kwargs)
response.status_code = 403
return response
This also allowed me to decide the status code based on some parameters of the request, which was necessary in my case to perform a 301 redirect for old-style URL patterns.
If you don't require the ability to change status code from within the method, then cjerdonek's answer is ideal.