Recieve url params with #api_view FBV - django-rest-framework

Is it possible to pass parameters from URL to a view function decorated with #api_view or I need to use APIView class instead?

Yes it's possible. What you've to do is access the request.query_params as below,
#api_view()
def sample_view(request, kw, *args,**kwargs):
url_params = request.query_params
# ypur code

Related

How can I disable authentication for GET method in django modelviewset?

I have written one router url and modelviewset like below:
**router.register(r'TestViewSet', views.TestViewSet)
**class TestViewSet(viewsets.ModelViewSet): queryset = Test.objects.all() serializer_class = TestSerializer
for this class, I want to disable authentication for GET method.
Pls give some solution
Well to disable authentication only on the GET method of the ModelViewSet, we are required to override the permission class based on the HTTP method.
To achieve this, try following
from rest_framework import permissions
...
class TestViewSet(viewsets.ModelViewSet):
...
def get_permissions(self):
"""Returns the permission based on the type of action"""
if self.action == "list":
return [permissions.AllowAny()]
return [permissions.IsAuthenticated()]
In the above code, we are checking of the action (assuming you want to allow anyone to see the list, and limit who can perform other actions).
So if the action is list (HTTP method GET) it will allow anyone to access otherwise it will check for authentication.
Hope this answers your question.

Django rest framework XLSX renderer + Apiview

I'm setting up an endpoint on my API which should return a XLSX-file. The DRF-docs mention https://github.com/wharton/drf-renderer-xlsx as the main external rendering package, aside from the pandas which also seem to be able to render XLSX.
In their code example they use a ReadOnlyViewset paired with a mixin, but there is no mention of how it's used with APIViews. Still, I would like to use an APIView as shown by this https://harshahegde.dev/rendering-xlsx-files-in-django-rest-framework-ckagk293p00eumks1bf4dlhie
However..
This works great when using CURL or Postman, but when done through a browser I get this error:
'Request' object has no attribute 'accepted_renderer'
From what I understand this is because there is no Accept header set (e.g 'Accept':'application/xlsx')
I fixed this by removing the Mixin from the renderer_classes, so it simply returns a file called "xlsx" but I can't figure out how to set the filename without the mixin. How do I set the filename using an APIView trying to access the URL from a browser?
My view:
class SomedataXlsx(APIView):
renderer_classes = [XLSXRenderer, JSONRenderer]
def get(self, request):
queryset = Somedata.objects.all()
serializer = SomeDataSerializer(queryset, many=True)
return Response(serializer.data)
Looking at the mixin code it became clear they change the content-disposition header, and so since the DRF Response() takes a header argument I tried changing it, and it worked perfectly.
class SomedataXlsx(APIView):
renderer_classes = [XLSXRenderer, JSONRenderer]
def get(self, request):
user_sub_fund_data = Somedata.objects.all()
serializer = SomeDataSerializer(queryset, many=True)
return Response(serializer.data, headers={"content-disposition":"attachment; filename=mynewfilename.xlsx"})

Parameters URL with DRF routers

I'm using Django Rest Framework for created a API. In this project i want to capture parameters in the URL. For example i want to capture the username and password of a user and my idea is like this:
http://localhost:8000/accounts/login/?unsername=username&password=password
But i cant, I' usin routers and django-filter, but i cant get the parameters url. My project files are there:
view.py:
class AccountsData(viewsets.ModelViewSet):
queryset = models.UserData.objects.all()
serializer_class = serializers.AccountsDataSerializer
permission_classes = (IsAuthenticated,)
filter_backends = (filters.DjangoFilterBackend,)
filterset_fields = ['username', 'password']
lookup_url_kwarg = 'username'
#action(methods=['get'], detail=True, url_name='login', url_path='login')
def login(self, request, pk=None):
return Response({"Login successfully"}, 200)
urls.py:
from api import views
router = routers.SimpleRouter()
router.register(r'accounts', views.AccountsData)
Request query parameters have nothing to do with routing, they are passed with the request independently of how you configure the route. You have access to them in request.query_params, for example, request.query_params.get('username') would get the value of the username parameter.
Being said that, your idea has a terrible mistake: password or any kind of confidential data should NEVER go in query parameters, you should use an http verb that carries the data in its body (POST, for example).

Django rest framework - exclude endpoint from authentication

I need to make view accessible without authentication, based on a variable passed in urls.py.
My idea is something like this:
urls.py
url(r'^oidc-api/', include('api.urls'), {'logged': True})
views.py
class ExampleViewSet(ModelViewSet):
if logged: # How can I get this variable, passed in urls.py?
permission_classes = () # This will exclude current view from authentication
queryset = Widget.objects.filter(visible=True)
serializer_class = ExampleSerializer
filter_backends = (DjangoFilterBackend,)
filter_fields = ('example_id',)
However, I can not access logged variable that is passed from urls.py.
John
You can't do that.
What you could do is, remove those permission classes from the views manually or set those to an empty set as,
class ExampleViewSet(ModelViewSet):
permission_classes = ()

How to access a page's URL in VoltRb

I'm trying to run some code on a controller in Volt, but only on certain pages. I've looked through the docs, but I'm not really sure how to get access to a given page's URL. Is there perhaps some hidden variable or something in the page model, like so?:
module Main
class MyController < Volt::ModelController
model :page
def index
end
def template_page
if page.url == "/foo/bar" # obviously, this doesn't actually work
# run some code
end
end
end
end
If you are using bindings in your URL, for example
client '/examples/{{ category }}/{{ example }}', controller: 'examples', action: 'template'
you can access those in the controller via the params collection:
params._category
params._example
In other cases you URL should be static anyway.
Sorry for the late reply. I added docs for the url method which is available from controllers:
http://docs.voltframework.com/en/docs/url.html

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