python - 在Django Rest框架中传递几个参数

Question

我是 Django 和 Django Rest Framework 的新手，我尝试创建多个路由来从数据库获取数据。

现在在我的 urls.py 文件中我有这个

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

这返回这个

"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.py我有这个

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)

现在我希望我的网址接受类似的内容

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

为此，我添加了这个新网址

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

但是现在当我尝试这个

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

我收到“找不到页面”

我知道这个URL必须调用检索函数来获取参数，那么，为什么会出现这个错误？

此 URL 不会执行与

相同的过程

http://127.0.0.1:8000/cpuProjects/ad 

提前致谢!

Answer 1

这与我们之前的 Q&A 中所做的有点相似

from rest_framework.decorators import detail_route, list_route
@detail_route(url_path='(?P<slug>[\w-]+)/(?P<what>[\w-]+)')
def get_by_name(self, request, pk=None,slug=None, what=None):
    print(slug, what)

同样，您可以对 list_route 执行相同的操作