Django Rest Framework,超链接嵌套关系

Question

我有两个模型:用户和票证.票证有一个用户,用户有很多票证

我已经完成了当我去url / users/1/ticket时,我得到了用户票的列表.

我想使用超链接关系,这是我在用户模型表示中看到的内容:

"tickets": [
    "http://127.0.0.1:8000/tickets/5/", 
    "http://127.0.0.1:8000/tickets/6/"
]

但我希望它像

"tickets": "http://127.0.0.1:8000/users/1/tickets"

有没有办法用DRF做到这一点？

网址:

    url(r'^users/(?P<user_pk>\d+)/tickets/$',
    views.TicketsByUserList.as_view(),
    name='myuser-tickets'),

风景:

class TicketsByUserList(generics.ListAPIView):
    model = Ticket
    serializer_class = TicketSerializer

    def get_queryset(self):
        user_pk = self.kwargs.get('user_pk', None)
        if user_pk is not None:
            return Ticket.objects.filter(user=user_pk)
        return []

用户序列化程序(我尝试使用故障单字段定义,更改类型,view_name,但没有效果):

class UserSerializer(serializers.HyperlinkedModelSerializer):
    tickets = serializers.HyperlinkedRelatedField(many=True, view_name='ticket-detail')

    class Meta:
        model = MyUser
        fields = ('id', 'nickname', 'email', 'tickets')

票证序列化器:

class TicketSerializer(serializers.HyperlinkedModelSerializer):
    user = serializers.HyperlinkedRelatedField(view_name='myuser-detail')
    liked = serializers.Field(source='liked')

    class Meta:
        model = Ticket
        fields = ('id', 'user', 'word', 'transcription', 'translation', 'liked', 'created', 'updated')

Answer 1

您可以使用SerializerMethodField来自定义它.像这样的东西:

class UserSerializer(serializers.HyperlinkedModelSerializer):
    tickets = serializers.SerializerMethodField('get_tickets')

    def get_tickets(self, obj):
        return "http://127.0.0.1:8000/users/%d/tickets" % obj.id

    class Meta:
        model = MyUser
        fields = ('id', 'nickname', 'email', 'tickets')

为了简洁起见,我在那里硬连接了URL,但你也可以进行反向查找.这基本上只是告诉它调用get_tickets方法而不是超类中的默认行为.