使用drf ysag记录django rest framewrok会生成“加载api定义失败”错误

yizd12fk  于 2021-09-08  发布在  Java
关注(0)|答案(0)|浏览(196)

我正在尝试记录我的django rest框架api,下面是输入链接描述文档。
我首先看到一些错误:pycharm在schema_viw.with_ui(..)上说“未解析的属性引用”with_ui'for class'type'”。下面是代码。

from django.contrib import admin
from django.urls import path , include
from django.conf import settings
from django.conf.urls.static import static
from rest_framework import permissions
from drf_yasg.views import get_schema_view
from drf_yasg import openapi

from rest_framework_swagger.views import get_swagger_view

schema_view = get_schema_view(
   openapi.Info(
      title="Snippets API",
      default_version='v1',
      description="Test description",
      terms_of_service="https://www.google.com/policies/terms/",
      contact=openapi.Contact(email="contact@snippets.local"),
      license=openapi.License(name="BSD License"),
   ),
   public=True,
   permission_classes=(permissions.AllowAny,),
)

urlpatterns = [
  path('swagger/', schema_view.with_ui('swagger', cache_timeout=0), name='schema-swagger-ui'),
  path('auth/', include('user_auth.urls')),
  path('api/', include('data_table.urls')),
] + static(settings.MEDIA_URL, document_root = settings.MEDIA_ROOT)

第二:我什么时候去http://127.0.0.1:8000/swagger/ 我看到加载api定义失败错误。附加的图片显示

确切的错误。我怎样才能解决这个错误。
以下是我的设置:

INSTALLED_APPS = [
    'django.contrib.admin' ,
    'django.contrib.auth' ,

    'django.contrib.contenttypes' ,
    'django.contrib.sessions' ,
    'django.contrib.messages' ,
    'django.contrib.staticfiles' ,
    'drf_yasg',
    'user_auth.apps.UserAuthConfig' ,
    'rest_framework_swagger',
    'rest_framework',
    'django_filters' ,
    'data_table',
    'rest_framework_simplejwt.token_blacklist',
    'corsheaders',
]

REST_FRAMEWORK = {
    # 'DEFAULT_PERMISSION_CLASSES': [
    #     'rest_framework.permissions.IsAuthenticated',
    # ],

    'DEFAULT_FILTER_BACKENDS': ['django_filters.rest_framework.DjangoFilterBackend'],
    'NON_FIELD_ERRORS_KEY' : 'error' ,

    # 'DEFAULT_AUTHENTICATION_CLASSES' : (
    #     'rest_framework_simplejwt.authentication.JWTAuthentication' ,
    #     )

暂无答案!

目前还没有任何答案,快来回答吧!

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