4

一个使用 Django、DRF、swagger 的项目。

网址配置如下：

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 {}

之后python manage.py runserver，访问http://127.0.0.1:8000/#并发生 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对象不能像 dict 一样赋值。

如果我将方法名称从detail_exportto重命名details_export，一切都会再次正常。

rest_framework 的@list_route装饰器将方法的 url 传递给 Link 对象时发生猜测错误。

为什么其他方法就好了？我该如何解决这个问题？

4

2 回答 2

3

这是DRF 中的错误。可能，它将在 3.6.4 中修复。现在你可以：

  1. 删除url_path自list_route：

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

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

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

  2. 或者使用带有自定义SchemaView的自定义SchemaGenerator：

    # 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()))
    # ...
]
于 2017-07-26T10:00:46.730 回答
1

这是一个黑客：

  1. 打开你的环境文件夹/lib/python3.5/site-packages/rest_framework/schemas.py
  2. 查找“insert_into”函数

  3. 替换这个：

    target[keys[-1]] = value

有了这个：

if type(target) != coreapi.Link:
    target[keys[-1]] = value
于 2017-07-10T22:44:57.317 回答