Python Boto - 将文件上传到 Amazon S3 上的特定位置
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Boto - Uploading file to a specific location on Amazon S3
提问by Jimmy
This is the code I'm working from
这是我正在使用的代码
import sys
import boto
import boto.s3
# AWS ACCESS DETAILS
AWS_ACCESS_KEY_ID = ''
AWS_SECRET_ACCESS_KEY = ''
bucket_name = AWS_ACCESS_KEY_ID.lower() + '-mah-bucket' conn = boto.connect_s3(AWS_ACCESS_KEY_ID, AWS_SECRET_ACCESS_KEY)
bucket = conn.create_bucket(bucket_name, location=boto.s3.connection.Location.DEFAULT)
uploadfile = sys.argv[1]
print 'Uploading %s to Amazon S3 bucket %s' % \
(uploadfile, bucket_name)
def percent_cb(complete, total):
sys.stdout.write('.')
sys.stdout.flush()
from boto.s3.key import Key
k = Key(bucket)
k.key = 'my test file'
k.set_contents_from_filename(testfile, cb=percent_cb, num_cb=10)
On my S3 I have created "directories", like this "bucket/images/holiday". I know these are only virtual directories.
在我的 S3 上，我创建了“目录”，例如“bucket/images/holiday”。我知道这些只是虚拟目录。
My question is, how can I modify this upload specifically to bucket/images/holiday virtual directory on S3 rather than the bucket root?
我的问题是，如何专门将此上传修改为 S3 上的存储桶/图像/假日虚拟目录而不是存储桶根目录？
采纳答案by garnaat
All you should have to do is prepend the virtual directory path to the key name prior to uploading. For example:
您所要做的就是在上传之前将虚拟目录路径添加到密钥名称之前。例如：
key_name = 'my test file'
path = 'images/holiday'
full_key_name = os.path.join(path, key_name)
k = bucket.new_key(full_key_name)
k.set_contents_from_filename(...)
You may have to change that a bit for your application but hopefully that gives you the basic idea.
您可能需要为您的应用程序稍微更改它，但希望这可以为您提供基本的想法。
回答by olekb
You can also use this:
你也可以使用这个：
from boto.s3.key import Key
bucket = conn.get_bucket('images')
k = Key(bucket)
k.key = 'holiday/my_test_file.txt'
It works like this in my s3_percent_uploader
它在我的s3_percent_uploader 中是这样工作的
python.exe s3_percent_uploader.py test_upload.txt testbucket test/upload.txt
after upload file will be in testbucket/test/upload.txt
上传文件后将在 testbucket/test/upload.txt
