Amazon S3 boto - how to create a folder?
How can I create a folder under a bucket using boto
library for Amazon s3?
I followed the manual, and created the key开发者_Python百科s with permission, metadata etc, but no where in the boto's documentation it describes how to create folders under a bucket, or create a folder under folders in bucket.
There is no concept of folders or directories in S3. You can create file names like "abc/xys/uvw/123.jpg"
, which many S3 access tools like S3Fox
show like a directory structure, but it's actually just a single file in a bucket.
Assume you wanna create folder abc/123/ in your bucket, it's a piece of cake with Boto
k = bucket.new_key('abc/123/')
k.set_contents_from_string('')
Or use the console
Use this:
import boto3
s3 = boto3.client('s3')
bucket_name = "YOUR-BUCKET-NAME"
directory_name = "DIRECTORY/THAT/YOU/WANT/TO/CREATE" #it's name of your folders
s3.put_object(Bucket=bucket_name, Key=(directory_name+'/'))
With AWS SDK .Net works perfectly, just add "/" at the end of the folder name string:
var folderKey = folderName + "/"; //end the folder name with "/"
AmazonS3 client = Amazon.AWSClientFactory.CreateAmazonS3Client(AWSAccessKey, AWSSecretKey);
var request = new PutObjectRequest();
request.WithBucketName(AWSBucket);
request.WithKey(folderKey);
request.WithContentBody(string.Empty);
S3Response response = client.PutObject(request);
Then refresh your AWS console, and you will see the folder
Tried many method above and adding forward slash /
to the end of key name, to create directory didn't work for me:
client.put_object(Bucket="foo-bucket", Key="test-folder/")
You have to supply Body
parameter in order to create directory:
client.put_object(Bucket='foo-bucket',Body='', Key='test-folder/')
Source: ryantuck in boto3 issue
Append "_$folder$" to your folder name and call put.
String extension = "_$folder$";
s3.putObject("MyBucket", "MyFolder"+ extension, new ByteArrayInputStream(new byte[0]), null);
see: http://www.snowgiraffe.com/tech/147/creating-folders-programmatically-with-amazon-s3s-api-putting-babies-in-buckets/
Update for 2019, if you want to create a folder with path bucket_name/folder1/folder2 you can use this code:
from boto3 import client, resource
class S3Helper:
def __init__(self):
self.client = client("s3")
self.s3 = resource('s3')
def create_folder(self, path):
path_arr = path.rstrip("/").split("/")
if len(path_arr) == 1:
return self.client.create_bucket(Bucket=path_arr[0])
parent = path_arr[0]
bucket = self.s3.Bucket(parent)
status = bucket.put_object(Key="/".join(path_arr[1:]) + "/")
return status
s3 = S3Helper()
s3.create_folder("bucket_name/folder1/folder2)
It's really easy to create folders. Actually it's just creating keys.
You can see my below code i was creating a folder with utc_time as name.
Do remember ends the key with '/' like below, this indicates it's a key:
Key='folder1/' + utc_time + '/'
client = boto3.client('s3')
utc_timestamp = time.time()
def lambda_handler(event, context):
UTC_FORMAT = '%Y%m%d'
utc_time = datetime.datetime.utcfromtimestamp(utc_timestamp)
utc_time = utc_time.strftime(UTC_FORMAT)
print 'start to create folder for => ' + utc_time
putResponse = client.put_object(Bucket='mybucketName',
Key='folder1/' + utc_time + '/')
print putResponse
Although you can create a folder by appending "/" to your folder_name. Under the hood, S3 maintains flat structure unlike your regular NFS.
var params = { Bucket : bucketName, Key : folderName + "/" }; s3.putObject(params, function (err, data) {});
S3 doesn't have a folder structure, But there is something called as keys.
We can create /2013/11/xyz.xls
and will be shown as folder's in the console. But the storage part of S3 will take that as the file name.
Even when retrieving we observe that we can see files in particular folder (or keys) by using the ListObjects
method and using the Prefix
parameter.
Apparently you can now create folders in S3. I'm not sure since when, but I have a bucket in "Standard" zone and can choose Create Folder from Action dropdown.
This question is more relevant to the future, so adding this update. I am using the upload_file method as shown below.
fold ='/my/system/filePath/tabmcq/Tables/auto/18.tsv'
s3_client.upload_file(
Filename = full/file/path/filename.extension,
Bucket = "tab-mcq-de",
Key = f"{fold.split('/')[-3]}/{fold.split('/')[-2]}/{fold.split('/')[-1]}"
)
Ideas is the "Filename" parameter requires the absolute file path of your system. The "Key" parameter requires the relative file path from the source directory where your files are located
In case of this example, Key parameter has to contain "Tables/auto/18.tsv" value, for client to create the folders.
Hope this helps.
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