【问题标题】:How to download file using django-storage sftp?如何使用 django-storage sftp 下载文件?
【发布时间】:2026-01-08 23:50:01
【问题描述】:

我正在使用 django 存储模块将文件存储在使用 sftp 的位置,现在我们需要从 sftp 下载相同的文件。

有什么建议吗?

models.py

from storages.backends.sftpstorage import SFTPStorage
SFS = SFTPStorage()

class Configurations(BaseModel):
name = models.CharField(max_length=150, unique=True)
file = models.FileField(upload_to='configurations', storage=SFS)
descriptions = models.TextField(null=True, blank=True)

settings.py

DEFAULT_FILE_STORAGE = 'storages.backends.sftpstorage.SFTPStorage'

SFTP_STORAGE_HOST = '127.0.0.1'
SFTP_STORAGE_ROOT = '/var/www/media/'
SFTP_STORAGE_PARAMS = {
    'username': 'root',
    'password': 'password',
    'allow_agent': False,
    'look_for_keys': False,
}
# SFTP_KNOWN_HOST_FILE = '~/.ssh/known_hosts'
SFTP_STORAGE_INTERACTIVE = False

views.py

from django.http import Http404

from rest_framework import viewsets
from rest_framework.parsers import MultiPartParser, FormParser

from .models import (CustomConfigurations)
from .serializers import (CustomConfigurationsSerializers)


class ToolCustomConfigurationsViewSet(viewsets.ModelViewSet):
    queryset = CustomConfigurations.objects.all()
    parser_classes = (FormParser, MultiPartParser,)
    serializer_class = CustomConfigurationsSerializers

    def get_queryset(self):
        queryset = self.queryset.all().order_by('-id')
        return queryset

serializer.py

class CustomConfigurationsSerializers(serializers.ModelSerializer):
class Meta:
    model = CustomConfigurations
    fields = '__all__'

【问题讨论】:

    标签: python django django-rest-framework sftp


    【解决方案1】:

    这是我的解决方案:

    @action(methods=['get'], detail=True)
    def download(self, request, pk=None):
    
        try:
            obj = CustomConfigurations.objects.get(id=pk)
        except CustomConfigurations.DoesNotExist:
            raise Http404
    
        if SFS.exists(obj.file.name):
            file = SFS._read(obj.file.name)
            type, encoding = mimetypes.guess_type(obj.file.name)
            response = HttpResponse(file, content_type=type)
            response['Content-Disposition'] = u'attachment; filename="{filename}'.format(
                filename=obj.file.name)
            return response
        raise Http404
    

    【讨论】: