在 Views.py 中,我有这些代码:
class ServerViewSet(viewsets.ViewSet):
def list(self, request):
servers = Server.objects.all()
serializer = ServerSerializer(servers, many=True)
return Response(serializer.data) # In here, I want to get the server name only
def retrieve(self, request, pk=None):
servers = get_object_or_404(Server, pk=pk)
serializer = ServerSerializer(servers)
return Response(serializer.data) # In here, I want to get the server name and ip address
在 serializers.py 中,我有这些代码:
class ServerSerializer(serializers.HyperlinkedModelSerializer):
class Meta:
model = Server
# fields = '__all__'
fields = ('id', 'name', 'desc', 'ip_address')
有没有官方的方法从serializers.data中过滤掉特定的字段
带有父类的示例
class CommonSerializer(serializers.ModelSerializer):
def get_field_names(self, *args, **kwargs):
field_names = self.context.get('fields', None)
if field_names:
return field_names
return super(self).get_field_names(*args, **kwargs)
class ServerSerializer(CommonSerializer):
class Meta:
..............
class WebsiteSerializer(CommonSerializer):
class Meta:
..............
最佳答案
在序列化程序中覆盖 get_field_names
:
class ServerSerializer(serializers.HyperlinkedModelSerializer):
def get_field_names(self, *args, **kwargs):
field_names = self.context.get('fields', None)
if field_names:
return field_names
return super(ServerSerializer, self).get_field_names(*args, **kwargs)
然后像这样使用它:
serializer = ServerSerializer(servers, context={'fields': ['name']})
serializer = ServerSerializer(servers, context={'fields': ['name', 'ip_address']})
* 编辑 *
将它变成 mixin,使其更易于重用:
class FieldMixin(object):
def get_field_names(self, *args, **kwargs):
field_names = self.context.get('fields', None)
if field_names:
return field_names
return super(FieldMixin, self).get_field_names(*args, **kwargs)
并使用它:
class ServerSerializer(FieldMixin, serializers.HyperlinkedModelSerializer):
关于python - 如何从 Django rest 框架的序列化程序中获取特定字段,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/47119879/