Question

我想设计一个API来处理我的django媒体目录中的所有文件以响应GET请求 - 我的应用程序的view.py应该是什么样的？

Answer 1

使用FileUploadParser解析原始文件上传内容。 request.data属性将是一个包含上传文件的单个“文件”的字典。

class FileUploadView(views.APIView):
    parser_classes = (FileUploadParser,)

    def put(self, request, filename, format=None):
        file_obj = request.FILES['file']
        save_location = open(MEDIA_ROOT + file_obj.name, 'wb+')
        for chunk in up_file.chunks():
            save_location.write(chunk)
            save_location.close()
        return Response(status=204)

Answer 2

功能基础代码

ufile = MODEL_NAME.objects.filter(user_id=user_id)
if ufile:
    uf = ufile[0]
    path = request.META['HTTP_HOST']
    path1 = "http://" + path + settings.MEDIA_URL
    url = path1 + str(uf.file).replace("./", "")
    dd = {'file': url}
    return Response(dd)
else:
    return Response(message="no file")

Answer 3

如果UR使用序列化器

#serializers.PY
class ProfileImageSerialiser(serializers.ModelSerializer):
    class Meta:
        model = MODEL_NAME
        fields = ('FIELD1', 'FIELD2')

#views.py
class GetProfileImageViewSet(generics.ListAPIView):
    serializer_class = ProfileImageSerialiser

    def get_queryset(self):
        user_id = self.kwargs['user_id']
        image = MODEL_NAME.objects.filter(user_id=user_id)
        return image

如何使用Django rest框架来提供Media目录？

3 个答案: