Python 在Django rest框架中传递几个参数

我对

Django

和

Django rest框架

是新手，我尝试创建几个路由来从数据库获取数据

现在在我的

url.py

文件中，我有一个

router = routers.DefaultRouter()
router.register(r'cpuProjects', cpuProjectsViewSet, base_name='cpuProjects'),

class cpuProjectsViewSet(viewsets.ViewSet):
  serializer_class = serializers.cpuProjectsSerializer
  # lookup_field = 'project_name'
  lookup_url_kwarg = 'project_name'

  def list(self, request):
    all_rows = connect_database()
    serializer = serializers.cpuProjectsSerializer(instance=all_rows, many=True)
    return Response(serializer.data)

  def retrieve(self, request, project_name=None):
    try:
      opc = {'name_proj' : project_name }
      all_rows = connect_database(opc)
    except KeyError:
        return Response(status=status.HTTP_404_NOT_FOUND)
    except ValueError:
        return Response(status=status.HTTP_400_BAD_REQUEST)
    serializer = serializers.cpuProjectsSerializer(instance=all_rows, many=True)
    return Response(serializer.data)

这个还这个

"cpuProjects": "http://127.0.0.1:8000/cpuProjects/"

我有可能做到这一点

http://127.0.0.1:8000/cpuProjects/

=>返回所有项目

http://127.0.0.1:8000/cpuProjects/ad

=>返回特定项目

在我看来，我有这个

router = routers.DefaultRouter()
router.register(r'cpuProjects', cpuProjectsViewSet, base_name='cpuProjects'),

class cpuProjectsViewSet(viewsets.ViewSet):
  serializer_class = serializers.cpuProjectsSerializer
  # lookup_field = 'project_name'
  lookup_url_kwarg = 'project_name'

  def list(self, request):
    all_rows = connect_database()
    serializer = serializers.cpuProjectsSerializer(instance=all_rows, many=True)
    return Response(serializer.data)

  def retrieve(self, request, project_name=None):
    try:
      opc = {'name_proj' : project_name }
      all_rows = connect_database(opc)
    except KeyError:
        return Response(status=status.HTTP_404_NOT_FOUND)
    except ValueError:
        return Response(status=status.HTTP_400_BAD_REQUEST)
    serializer = serializers.cpuProjectsSerializer(instance=all_rows, many=True)
    return Response(serializer.data)

现在我希望我的Url接受这样的内容

http://127.0.0.1:8000/cpuProjects/ad/comments

http://127.0.0.1:8000/cpuProjects/ad/ussing

http://127.0.0.1:8000/cpuProjects/ad/process

为此，我添加了这个新的url

router.register(r'cpuProjects/([a-zA-Z0-9]+)$', cpuProjectsViewSet, base_name='cpuProjects'),

但是现在当我尝试这个的时候

http://127.0.0.1:8000/cpuProjects/ad/ussing

我获得“未找到页面”

我知道这个URL必须调用检索函数来获取参数，所以，为什么会出现这个错误

此URL不会执行与以下相同的过程

http://127.0.0.1:8000/cpuProjects/ad 

---

提前谢谢

路由器框架不适用于多个参数。。您可以使用主键（在正则表达式中）手动执行此操作。

这与我们在前面的

从rest\u framework.decorators导入细节\u路由、列表\u路由
@详细信息路径（url路径='（？P[\w-]+）/（？P[\w-]+））
def按名称获取（self、request、pk=None、slug=None、what=None）：
打印（slug，什么）

同样，您也可以对

列表\u路线执行相同的操作