如何阻止drf spectacular自动生成无需展示的接口Schema？

• 单个视图/动作排除：给目标视图或视图动作加上@extend_schema(exclude=True)装饰器，直接标记不纳入Schema生成。
  单个视图动作示例：

  from drf_spectacular.utils import extend_schema
  
  class UnwantedAPIView(APIView):
      @extend_schema(exclude=True)
      def get(self, request):
          return Response({"data": "内容"})
  

  整个视图集排除示例：

  @extend_schema(exclude=True)
  class UnwantedViewSet(viewsets.ModelViewSet):
      queryset = SomeModel.objects.all()
      serializer_class = SomeSerializer
  

• 全局URL批量过滤：在settings.py的SPECTACULAR_SETTINGS里，用SERVE_EXCLUDE配置要排除的URL正则，批量过滤接口：

  SPECTACULAR_SETTINGS = {
      # 保留原有其他配置
      'SERVE_EXCLUDE': [
          r'^internal-api/.*$',
          r'^test/.*$',
      ],
  }
  

• 路由层面直接排除：定义URL时给不需要的接口加上include_in_schema=False参数，从路由源头上跳过Schema生成：

  urlpatterns = [
      path('unwanted/', UnwantedAPIView.as_view(), name='unwanted', include_in_schema=False),
      # 正常接口保留配置
      path('normal/', NormalAPIView.as_view(), name='normal'),
  ]
  

• 权限控制自动排除：如果是内部专用接口，自定义权限类，让Schema生成时自动跳过这类接口：

  from rest_framework.permissions import BasePermission
  
  class InternalOnly(BasePermission):
      def has_permission(self, request, view):
          # 生成Schema的请求会携带swagger_fake_view标记，直接返回False排除
          if getattr(request, 'swagger_fake_view', False):
              return False
          return request.user.is_staff
  
  class InternalAPIView(APIView):
      permission_classes = [InternalOnly]
  

内容的提问来源于stack exchange，提问作者knowNothing