1个模型的django-rest-framework多个序列化器？

Question

假设你想放弃

{field1, field2, field3} 详细请求。
{field1, field2} 列表请求。
{field1} 其他更简单的列表请求。

我已经看到使用 get_serializer_class 和 self.action 的示例，它可以处理细节与列表的情况。 (https://stackoverflow.com/a/22755648/433570)

我应该定义两个视图集和两个 url 端点吗？
还是这里有更好的方法？

我想这也可以应用于美味派。 （两个资源？）

Answer 1

我自己没有测试过，但我认为你可以覆盖你需要的方法。

根据the documentation（标记路由的额外操作），您可以这样做：

class UserViewSet(viewsets.ViewSet):
"""
Example empty viewset demonstrating the standard
actions that will be handled by a router class.

If you're using format suffixes, make sure to also include
the `format=None` keyword argument for each action.
"""

def list(self, request):
    pass

def create(self, request):
    pass

def retrieve(self, request, pk=None):
    pass

def update(self, request, pk=None):
    pass

def partial_update(self, request, pk=None):
    pass

def destroy(self, request, pk=None):
    pass

或者如果您需要自定义方法：

from django.contrib.auth.models import User
from rest_framework import status
from rest_framework import viewsets
from rest_framework.decorators import detail_route, list_route
from rest_framework.response import Response
from myapp.serializers import UserSerializer, PasswordSerializer

class UserViewSet(viewsets.ModelViewSet):
"""
A viewset that provides the standard actions
"""
queryset = User.objects.all()
serializer_class = UserSerializer

@detail_route(methods=['post'])
def set_password(self, request, pk=None):
    user = self.get_object()
    serializer = PasswordSerializer(data=request.DATA)
    if serializer.is_valid():
        user.set_password(serializer.data['password'])
        user.save()
        return Response({'status': 'password set'})
    else:
        return Response(serializer.errors,
                        status=status.HTTP_400_BAD_REQUEST)

@list_route()
def recent_users(self, request):
    recent_users = User.objects.all().order('-last_login')
    page = self.paginate_queryset(recent_users)
    serializer = self.get_pagination_serializer(page)
    return Response(serializer.data)