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[英]check if POST request exist in database and return custom response in django rest framework
[英]Return custom response to successful POST request in django rest framework
我想在用戶使用 POST 請求訪問 API 時向用戶返回自定義響應,並且這是成功的。 以下是代碼片段: views.py
class BlogPostAPIView(mixins.CreateModelMixin,generics.ListAPIView):
# lookup_field = 'pk'
serializer_class = BlogPostSerializer
def get_queryset(self):
return BlogPost.objects.all()
def perform_create(self, serializer):
serializer.save(user=self.request.user)
def post(self,request,*args,**kwargs):
return self.create(request,*args,**kwargs)
網址.py
app_name = 'postings'
urlpatterns = [
re_path('^$', BlogPostAPIView.as_view(),name='post-create'),
re_path('^(?P<pk>\d+)/$', BlogPostRudView.as_view(),name='post-rud'),
]
現在它將發布請求的詳細信息作為成功響應返回,有什么方法可以根據我自己的自定義查詢集返回一些其他響應?
您可以在 views.py 上編寫自定義 api。 我想例如;
from rest_framework.views import APIView
from rest_framework.response import Response
class Hello(APIView):
@csrf_exempt
def post(self, request):
content = "Hi"
type = "message"
return Reponse({"content":content,"type":type})
然后定義網址。
app_name = 'postings'
urlpatterns = [
re_path('^$', BlogPostAPIView.as_view(),name='post-create'),
re_path('^(?P<pk>\d+)/$', BlogPostRudView.as_view(),name='post-rud'),
re_path('^hello/$', Hello.as_view(),name='Hello'),
]
就是這樣。
您還可以管理permessions: https://www.django-rest-framework.org/api-guide/permissions/#setting-the-permission-policy ,你可以使用的意見串行: HTTPS://www.django- rest-framework.org/api-guide/serializers/#saving-instances
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