python - django_rest_framwork 发生 "TypeError: ' 链接对象不支持项目分配”

Question

一个使用 Django、DRF、swagger 的项目。

url 配置如下:

schema_view = get_swagger_view(title='Pastebin API')
urlpatterns = [
    url(r'^$', schema_view),
    url(r'store', include('store.urls')),
    ... # other urls using routers.SimplerRouter
]

和存储/urls.py:

router = routers.SimpleRouter()
router.register(r'', views.StoreViewSet)
urlpatterns = router.urls

和views.StoreViewSet:

class StoreViewSet(BaseObject, GenericViewSet):
    permition_class = ()

    @list_route()
    def detail(self, request):
        return {}

    @list_route(url_path='detail/export')
    def detail_export(self, request):
        return {}

python manage.py runserver后，访问http://127.0.0.1:8000/#，出现TypeError:

File "/usr/local/share/.virtualenvs/dev-finance/lib/python2.7/site-packages/rest_framework_swagger/views.py", line 32, in get
schema = generator.get_schema(request=request)
File "/usr/local/share/.virtualenvs/dev-finance/lib/python2.7/site-packages/rest_framework/schemas.py", line 242, in get_schema
links = self.get_links(request)
File "/usr/local/share/.virtualenvs/dev-finance/lib/python2.7/site-packages/rest_framework/schemas.py", line 276, in get_links
insert_into(links, keys, link)
File "/usr/local/share/.virtualenvs/dev-finance/lib/python2.7/site-packages/rest_framework/schemas.py", line 79, in insert_into
target[keys[-1]] = value
TypeError: 'Link' object does not support item assignment
[ERROR] 2017-05-04 15:25:06,936 log_message(basehttp.py:131) "GET / HTTP/1.1" 500 20384

错误消息显示，Link 对象无法像 dict 那样赋值。

如果我将方法名称从 detail_export 重命名为 details_export，一切都会恢复正常。

rest_framework的@list_route装饰器出现猜测错误 将方法的 url 传输到 Link 对象。

为什么其他方法都可以？ 我该如何解决这个问题？

Answer 1

这是 the bug in DRF .可能会在 3.6.4 中修复。现在您可以:

1. 从 list_route 中删除 url_path:

   class StoreViewSet(BaseObject, GenericViewSet):
       permition_class = ()
   
       @list_route()
       def detail(self, request):
           return {}
   
       @list_route()
       def detail_export(self, request):
           return {}
   

2. 或者将自定义 SchemaView 与自定义 SchemaGenerator 一起使用:

   # api/schema.py for example
   
   from rest_framework.permissions import AllowAny
   from rest_framework.renderers import CoreJSONRenderer
   from rest_framework.response import Response
   from rest_framework.schemas import SchemaGenerator
   from rest_framework.views import APIView
   from rest_framework_swagger import renderers
   
   
   class CustomSchemaGenerator(SchemaGenerator):
       def get_keys(self, subpath, method, view):
           result = super().get_keys(subpath, method, view)
           # here you can fix your path
           if result == ['store', 'detail', 'detail_export']:  
               return ['store', 'detail_export'] 
           return result
   
   
   class SwaggerSchemaView(APIView):
       _ignore_model_permissions = True
       exclude_from_schema = True
       permission_classes = [AllowAny]
       renderer_classes = [
           CoreJSONRenderer,
           renderers.OpenAPIRenderer,
           renderers.SwaggerUIRenderer
       ]
   
       def get(self, request):
           generator = CustomSchemaGenerator(title='API')
           return Response(generator.get_schema(request=request))
   
   # urls.py
   
   from django.contrib.auth.decorators import login_required
   from api.schema import SwaggerSchemaView   
   
   urlpatterns = [
       # ...
       url(r'^swagger-url/', login_required(SwaggerSchemaView.as_view()))
       # ...
   ]