将上传的图像存储到AWS S3

Question

我有一个用python/django（REST api）编写的服务器应用程序，用于接受来自客户机应用程序的文件上传。我希望这个上传的文件存储在AWS S3中。我也希望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将帮助我理解应该如何完成。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() 

在此先感谢

Answer 1

看看boto包，它提供了AWS的API：

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() 

Answer 2

如果你想你的上传直接去AWS S3，您可以使用django-storages和将您的Django文件存储后端设置为使用AWS S3。

• django-storages
• django-storages documentation

这将使你的Django项目，以透明地处理存储于S3，而你不必手动您上传的文件重新上传到S3。

存储设置

您将需要至少这些配置添加到您的Django设置：

# 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 

示例代码上传存储到远程存储

这是一个您使用Django存储后端的handle_uploaded_file方法将视图的修改版本保存到远程文件stination（使用django-storages）。

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

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() 

参见如何访问这里的远程存储更多的解释和例子：

• django-storages: Amazon S3