如何使用 Django Rest 框架返回 HttpResponse？

Question

我正在构建一个 API 函数，允许客户端发送带有 URL 参数的 GET 请求，服务器根据给定的信息接收和处理文件，并返回一个自定义文件。好消息是我的所有步骤都是独立工作的！

我能够在返回 HttpResponse 的 def get_query 函数 except 中让一切正常工作。该函数需要一个 Queryset 响应（我猜是有道理的..）。我想我需要另一个函数来返回 HttpResponse，所以我创建了def send_file。我不知道如何调用这个函数，现在它只是跳过它。

views.py.

class SubFileViewSet(viewsets.ModelViewSet):

    queryset = subfiles.objects.all()
    serializer_class = SubFilesSerializer
    permission_classes = (permissions.IsAuthenticatedOrReadOnly,
                          IsOwnerOrReadOnly,)

    def send_file(self, request):

        req = self.request
        make = req.query_params.get('make')
        model = req.query_params.get('model')
        plastic = req.query_params.get('plastic')
        quality = req.query_params.get('qual')
        filenum = req.query_params.get('fileid')
        hotend = req.query_params.get('hotendtemp')
        bed = req.query_params.get('bedtemp')

        if make and model and plastic and quality and filenum:            

            filepath = subfiles.objects.values('STL', 'FileTitle').get(fileid = filenum)

            path = filepath['STL']
            title =  filepath['FileTitle']    

            '''
            #Live processing (very slow)
            gcode = TheMagic(
                make=make, 
                model=model, 
                plastic=plastic, 
                qual=quality, 
                path=path, 
                title=title, 
                hotendtemp=hotend,
                bedtemp=bed)

            '''
            #Local data for faster testing
            gcode = "/home/bradman/Documents/Programming/DjangoWebProjects/3dprinceprod/fullprince/media/uploads"

            test_file = open(gcode, 'r')
            response = HttpResponse(test_file, content_type='text/plain')
            response['Content-Disposition'] = "attachment; filename=%s" % title


            print (response)
            return response

    def perform_create(self, serializer):
        serializer.save(owner=self.request.user)        


    def get_queryset(self):
        req = self.request
        make = req.query_params.get('make')
        model = req.query_params.get('model')
        plastic = req.query_params.get('plastic')
        quality = req.query_params.get('qual')
        filenum = req.query_params.get('fileid')
        hotend = req.query_params.get('hotendtemp')
        bed = req.query_params.get('bedtemp')

        return self.queryset
        #function proved in here then removed and placed above
        #get_query required a queryset response

我对 Django Rest Framework 几乎没有经验，我不确定是否有更好的方法来实现它，或者我如何调用函数def send_file？

Answer 1

您正在寻找自定义路线。将您的send_file 设置为list_route http://www.django-rest-framework.org/api-guide/viewsets/#marking-extra-actions-for-routing

from rest_framework.decorators import list_route

class SubFileViewSet(viewsets.ModelViewSet):
    ...

    @list_route(methods=['get'])
    def send_file(self, request):
        ...

然后你就可以访问这个方法了

/subfile/send_file/?params