使用ModelViewSet在URL中传递自定义param

Question

使用传递的最佳方法是什么？例如，实现这样的目标：

http://127.0.0.1:8000/api/v1/financing-settings/template/?param=block

下面是我使用的方法，但是我发现我已经在身体部分设置了param，但这不是我想要的：

class TemplateView(ModelViewSet):
    """ViewSet for Saving Block/ Step template."""

   
    def list(self, request, *args, **kwargs):

        """Get list of Block/Steps with is_process_template is equal to True."""
        param = request.data['param']

        if param == "block":
            _block = Block.objects.filter(is_process_template=True).values()
            return JsonResponse({"data": list(_block)}, safe=False, status=200)

        elif param == "step":
            _step = Step.objects.filter(is_process_template=True).values()
            return JsonResponse({"data": list(_step)}, safe=False, status=200)

        return Response(status=status.HTTP_204_NO_CONTENT)

Answer 1

 param = request.GET.get('param')

或申请邮寄

 param = request.POST.get('param')

Answer 2

param =request.query_params.get(“param”)