Question

当我去http://127.0.0.1:8000/movies/时它得到错误：djangorestframework :type object has no attribute 'id'
为什么我无法获得id值？请帮帮我，谢谢！

这是我的代码

serialize.py

class MovieSerializer(serializers.ModelSerializer): 
    link = serializers.HyperlinkedRelatedField(source='id',  view_name='movie_detail')
    class Meta:
        model = Movie
        fields = ( 'link', 'title')


class MovieDetail(object):
    queryset = Movie.objects.all()
    serializer_class = MovieSerializer

urls.py

urlpatterns = patterns('',
    url(r'^movies/$', MovieList.as_view(), name='movie_list'),
    url(r'^movies/(?P<pk>[0-9]+)$', MovieDetail.as_view(), name='movie_detail'),

如何编辑它以显示如下结果：

"results": [
        { 
            "link": "http://127.0.0.1:8000/movies/1", 
            "title": "test", 
        },   
        { 
            "link": "http://127.0.0.1:8000/movies/2", 
            "title": "test", 
        },   
        ]

Answer 1

HyperlinkedRelatedField必须有source指向模型上的相关对象。您正在寻找HyperlinkedIdentityField，它会为当前对象提供一个链接，类似于HyperlinkedModelSerializer将自动为您执行的操作。

class MovieSerializer(serializers.ModelSerializer): 
    link = serializers.HyperlinkedIdentityField(view_name='movie_detail')

    class Meta:
        model = Movie
        fields = ( 'link', 'title', )

djangorestframework：type对象没有属性'id'

1 个答案: