Question

我有一个用python / django（REST api）编写的服务器应用程序，用于接受来自客户端应用程序的文件上载。我希望这个上传的文件存储在AWS S3中。我还希望将文件作为多部分表单/数据从客户端上传。我怎样才能做到这一点。任何示例代码应用程序都将帮助我理解它应该如何完成。请协助。

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

    def put(self, request, filename, format=None):
        file_obj = request.data['file']
        self.handle_uploaded_file(file_obj)
        return self.get_response("", True, "", {})

    def handle_uploaded_file(self, f):
        destination = open('<path>', 'wb+')
        for chunk in f.chunks():
            destination.write(chunk)
        destination.close()

提前致谢

Answer 1

如果您希望上传直接转到AWS S3，可以使用django-storages并设置Django文件存储后端以使用AWS S3。

这将允许您的Django项目透明地处理存储到S3，而无需手动将上传的文件重新上传到S3。

存储设置

您需要在Django设置中添加至少这些配置：

# default remote file storage
DEFAULT_FILE_STORAGE = 'storages.backends.s3boto.S3BotoStorage'

# aws access keys
AWS_ACCESS_KEY_ID = 'YOUR-ACCESS-KEY'
AWS_SECRET_ACCESS_KEY = 'YOUR-SECRET-ACCESS-KEY'
AWS_BUCKET_NAME = 'your-bucket-name'
AWS_STORAGE_BUCKET_NAME = AWS_BUCKET_NAME

存储上传到远程存储的示例代码

这是视图的修改版本，使用Django的存储后端handle_uploaded_file方法将uploade文件保存到远程目标（使用django-storages）。

注意：请务必在DEFAULT_FILE_STORAGE中定义settings和AWS密钥，以便django-storage可以访问您的存储分区。

from django.core.files.storage import default_storage
from django.core.files import File

# set file i/o chunk size to maximize throughput
FILE_IO_CHUNK_SIZE = 128 * 2**10


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

    def put(self, request, filename, format=None):
        file_obj = request.data['file']
        self.handle_uploaded_file(file_obj)
        return self.get_response("", True, "", {})

    def handle_uploaded_file(self, f):
        """
        Write uploaded file to destination using default storage.
        """
        # set storage object to use Django's default storage
        storage = default_storage

        # set the relative path inside your bucket where you want the upload
        # to end up
        fkey = 'sub-path-in-your-bucket-to-store-the-file'

        # determine mime type -- you may want to parse the upload header
        # to find out the exact MIME type of the upload file.
        content_type = 'image/jpeg'

        # write file to remote server
        # * "file" is a File storage object that will use your
        # storage backend (in this case, remote storage to AWS S3)
        # * "media" is a File object created with your upload file
        file = storage.open(fkey, 'w')
        storage.headers.update({"Content-Type": content_type})
        f = open(path, 'rb')
        media = File(f)
        for chunk in media.chunks(chunk_size=FILE_IO_CHUNK_SIZE):
            file.write(chunk)
        file.close()
        media.close()
        f.close()

在此处查看有关如何访问远程存储的更多说明和示例：

django-storages: Amazon S3

Answer 2

查看提供AWS API的boto包：

from boto.s3.connection import S3Connection
s3 = S3Connection(access_key, secret_key)
b = s3.get_bucket('<bucket>')
mp = b.initiate_multipart_upload('<object>')
for i in range(1, <parts>+1):
    io = <receive-image-part>   # E.g. StringIO
    mp.upload_part_from_file(io, part_num=i)
mp.complete_upload()

将上传的图像存储到AWS S3

2 个答案: