Key (Str) -- the Name of the Key to Upload to.
How to upload a file to directory in S3 bucket using boto
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- How to upload a file to directory in S3 bucket using boto
I want to copy a file in s3 bucket using python.
Ex : I have saucepan name = test. And in the bucket, I have two folders name "dump" & "input". Now I want to re-create a file from local directory to S3 "dump" folder using python... Can anyone aid me?
This question is tagged with python amazon-web-services amazon-s3 boto
~ Asked on 2013-02-26 09:47:53
fifteen Answers
Try this...
import boto import boto.s3 import sys from boto.s3.central import Central AWS_ACCESS_KEY_ID = '' AWS_SECRET_ACCESS_KEY = '' bucket_name = AWS_ACCESS_KEY_ID.lower() + '-dump' conn = boto.connect_s3(AWS_ACCESS_KEY_ID, AWS_SECRET_ACCESS_KEY) bucket = conn.create_bucket(bucket_name, location=boto.s3.connection.Location.DEFAULT) testfile = "replace this with an actual filename" impress 'Uploading %southward to Amazon S3 saucepan %due south' % \ (testfile, bucket_name) def percent_cb(complete, total): sys.stdout.write('.') sys.stdout.flush() grand = Key(bucket) k.key = 'my exam file' thou.set_contents_from_filename(testfile, cb=percent_cb, num_cb=10) [UPDATE] I am non a pythonist, so thank you for the heads up about the import statements. Too, I'd not recommend placing credentials inside your own source lawmaking. If you are running this inside AWS utilise IAM Credentials with Instance Profiles (http://docs.aws.amazon.com/IAM/latest/UserGuide/id_roles_use_switch-office-ec2_instance-profiles.html), and to keep the same behaviour in your Dev/Test environment, use something similar Hologram from AdRoll (https://github.com/AdRoll/hologram)
~ Answered on 2013-02-26 11:04:56
import boto3 s3 = boto3.resource('s3') BUCKET = "test" s3.Bucket(Saucepan).upload_file("your/local/file", "dump/file") ~ Answered on 2017-eleven-03 15:17:29
No need to make it that complicated:
s3_connection = boto.connect_s3() bucket = s3_connection.get_bucket('your bucket name') key = boto.s3.key.Key(bucket, 'some_file.zippo') with open('some_file.zip') as f: central.send_file(f) ~ Answered on 2015-06-29 10:00:eighteen
I used this and it is very simple to implement
import tinys3 conn = tinys3.Connexion('S3_ACCESS_KEY','S3_SECRET_KEY',tls=True) f = open('some_file.cipher','rb') conn.upload('some_file.zip',f,'my_bucket') https://world wide web.smore.com/labs/tinys3/
~ Answered on 2014-12-24 08:48:46
Upload file to s3 within a session with credentials.
import boto3 session = boto3.Session( aws_access_key_id='AWS_ACCESS_KEY_ID', aws_secret_access_key='AWS_SECRET_ACCESS_KEY', ) s3 = session.resources('s3') # Filename - File to upload # Saucepan - Bucket to upload to (the height level directory under AWS S3) # Primal - S3 object name (can comprise subdirectories). If not specified then file_name is used s3.meta.client.upload_file(Filename='input_file_path', Bucket='bucket_name', Primal='s3_output_key') ~ Answered on 2019-02-08 xv:22:27
from boto3.s3.transfer import S3Transfer import boto3 #have all the variables populated which are required below client = boto3.client('s3', aws_access_key_id=access_key,aws_secret_access_key=secret_key) transfer = S3Transfer(customer) transfer.upload_file(filepath, bucket_name, folder_name+"/"+filename) ~ Answered on 2017-01-31 12:34:05
This will also work:
import os import boto import boto.s3.connexion from boto.s3.key import Cardinal try: conn = boto.s3.connect_to_region('us-east-1', aws_access_key_id = 'AWS-Access-Central', aws_secret_access_key = 'AWS-Secrete-Key', # host = 's3-website-united states of america-east-1.amazonaws.com', # is_secure=Truthful, # uncomment if you are not using ssl calling_format = boto.s3.connectedness.OrdinaryCallingFormat(), ) bucket = conn.get_bucket('YourBucketName') key_name = 'FileToUpload' path = 'images/holiday' #Directory Nether which file should get upload full_key_name = os.path.join(path, key_name) k = bucket.new_key(full_key_name) k.set_contents_from_filename(key_name) except Exception,e: print str(eastward) impress "fault" ~ Answered on 2017-02-01 08:xix:25
This is a 3 liner. Just follow the instructions on the boto3 documentation.
import boto3 s3 = boto3.resource(service_name = 's3') s3.meta.client.upload_file(Filename = 'C:/foo/bar/baz.filetype', Bucket = 'yourbucketname', Key = 'baz.filetype') Some important arguments are:
Parameters:
str) -- The path to the file to upload.str) -- The name of the bucket to upload to. str) -- The proper noun of the that you want to assign to your file in your s3 saucepan. This could exist the same as the name of the file or a different name of your choice but the filetype should remain the same. Note: I assume that yous have saved your credentials in a ~\.aws binder every bit suggested in the best configuration practices in the boto3 documentation.
~ Answered on 2018-11-29 00:20:09
import boto from boto.s3.central import Fundamental AWS_ACCESS_KEY_ID = '' AWS_SECRET_ACCESS_KEY = '' END_POINT = '' # eg. us-east-ane S3_HOST = '' # eg. s3.us-east-i.amazonaws.com BUCKET_NAME = 'exam' FILENAME = 'upload.txt' UPLOADED_FILENAME = 'dumps/upload.txt' # include folders in file path. If it doesn't be, information technology volition be created s3 = boto.s3.connect_to_region(END_POINT, aws_access_key_id=AWS_ACCESS_KEY_ID, aws_secret_access_key=AWS_SECRET_ACCESS_KEY, host=S3_HOST) bucket = s3.get_bucket(BUCKET_NAME) yard = Cardinal(bucket) k.primal = UPLOADED_FILENAME chiliad.set_contents_from_filename(FILENAME) ~ Answered on 2017-03-02 13:23:42
Using boto3
import logging import boto3 from botocore.exceptions import ClientError def upload_file(file_name, bucket, object_name=None): """Upload a file to an S3 bucket :param file_name: File to upload :param bucket: Saucepan to upload to :param object_name: S3 object proper noun. If not specified then file_name is used :return: Truthful if file was uploaded, else Fake """ # If S3 object_name was not specified, apply file_name if object_name is None: object_name = file_name # Upload the file s3_client = boto3.client('s3') endeavor: response = s3_client.upload_file(file_name, bucket, object_name) except ClientError equally eastward: logging.error(eastward) return False return True For more:- https://boto3.amazonaws.com/v1/documentation/api/latest/guide/s3-uploading-files.html
~ Answered on 2019-08-22 11:10:50
For upload folder example equally post-obit code and S3 folder picture 
import boto import boto.s3 import boto.s3.connection import os.path import sys # Fill in info on data to upload # destination bucket proper name bucket_name = 'willie20181121' # source directory sourceDir = '/home/willie/Desktop/x/' #Linux Path # destination directory proper noun (on s3) destDir = '/test1/' #S3 Path #max size in bytes before uploading in parts. between 1 and 5 GB recommended MAX_SIZE = twenty * m * 1000 #size of parts when uploading in parts PART_SIZE = half-dozen * 1000 * 1000 access_key = 'MPBVAQ*******Information technology****' secret_key = '11t63yDV***********HgUcgMOSN*****' conn = boto.connect_s3( aws_access_key_id = access_key, aws_secret_access_key = secret_key, host = '******.org.tw', is_secure=False, # uncomment if you are not using ssl calling_format = boto.s3.connection.OrdinaryCallingFormat(), ) bucket = conn.create_bucket(bucket_name, location=boto.s3.connexion.Location.DEFAULT) uploadFileNames = [] for (sourceDir, dirname, filename) in os.walk(sourceDir): uploadFileNames.extend(filename) break def percent_cb(complete, full): sys.stdout.write('.') sys.stdout.flush() for filename in uploadFileNames: sourcepath = os.path.join(sourceDir + filename) destpath = os.path.join(destDir, filename) print ('Uploading %s to Amazon S3 saucepan %due south' % \ (sourcepath, bucket_name)) filesize = bone.path.getsize(sourcepath) if filesize > MAX_SIZE: print ("multipart upload") mp = saucepan.initiate_multipart_upload(destpath) fp = open up(sourcepath,'rb') fp_num = 0 while (fp.tell() < filesize): fp_num += 1 print ("uploading part %i" %fp_num) mp.upload_part_from_file(fp, fp_num, cb=percent_cb, num_cb=ten, size=PART_SIZE) mp.complete_upload() else: impress ("singlepart upload") chiliad = boto.s3.key.Key(bucket) m.key = destpath chiliad.set_contents_from_filename(sourcepath, cb=percent_cb, num_cb=10) PS: For more than reference URL
~ Answered on 2018-12-03 08:23:35
I have something that seems to me has a bit more club:
import boto3 from pprint import pprint from botocore.exceptions import NoCredentialsError class S3(object): BUCKET = "test" connection = None def __init__(cocky): endeavor: vars = get_s3_credentials("aws") self.connection = boto3.resource('s3', 'aws_access_key_id', 'aws_secret_access_key') except(Exception) equally mistake: impress(mistake) self.connection = None def upload_file(self, file_to_upload_path, file_name): if file_to_upload is None or file_name is None: return False try: pprint(file_to_upload) file_name = "your-binder-inside-s3/{0}".format(file_name) cocky.connexion.Saucepan(self.BUCKET).upload_file(file_to_upload_path, file_name) print("Upload Successful") return True except FileNotFoundError: impress("The file was non plant") return Fake except NoCredentialsError: print("Credentials not available") return False There're iii important variables here, the BUCKET const, the file_to_upload and the file_name
Saucepan: is the name of your S3 bucket
file_to_upload_path: must be the path from file yous want to upload
file_name: is the resulting file and path in your bucket (this is where you add folders or what always)
There'due south many ways merely y'all tin reuse this code in another script like this
import S3 def some_function(): S3.S3().upload_file(path_to_file, final_file_name) ~ Answered on 2020-06-29 23:31:09
You should mention the content blazon as well to omit the file accessing issue.
import bone epitome='fly.png' s3_filestore_path = 'images/fly.png' filename, file_extension = os.path.splitext(image) content_type_dict={".png":"image/png",".html":"text/html", ".css":"text/css",".js":"application/javascript", ".jpg":"paradigm/png",".gif":"paradigm/gif", ".jpeg":"epitome/jpeg"} content_type=content_type_dict[file_extension] s3 = boto3.customer('s3', config=boto3.session.Config(signature_version='s3v4'), region_name='ap-south-i', aws_access_key_id=S3_KEY, aws_secret_access_key=S3_SECRET) s3.put_object(Torso=image, Bucket=S3_BUCKET, Key=s3_filestore_path, ContentType=content_type) ~ Answered on 2020-ten-01 06:56:57
If you have the aws command line interface installed on your system you can make use of pythons subprocess library. For example:
import subprocess def copy_file_to_s3(source: str, target: str, bucket: str): subprocess.run(["aws", "s3" , "cp", source, f"s3://{bucket}/{target}"]) Similarly you lot can utilize that logics for all sort of AWS customer operations like downloading or listing files etc. It is besides possible to get return values. This way at that place is no demand to import boto3. I judge its apply is non intended that manner but in practise I observe it quite convenient that way. This way you also become the status of the upload displayed in your panel - for example:
Completed iii.5 GiB/3.5 GiB (242.8 MiB/s) with ane file(s) remaining To modify the method to your wishes I recommend having a look into the subprocess reference as well as to the AWS Cli reference.
Note: This is a copy of my reply to a similar question.
~ Answered on 2020-xi-06 11:41:forty
xmlstr = etree.tostring(listings, encoding='utf8', method='xml') conn = boto.connect_s3( aws_access_key_id = access_key, aws_secret_access_key = secret_key, # host = '<bucketName>.s3.amazonaws.com', host = 'bycket.s3.amazonaws.com', #is_secure=Faux, # uncomment if you are not using ssl calling_format = boto.s3.connection.OrdinaryCallingFormat(), ) conn.auth_region_name = 'us-west-1' bucket = conn.get_bucket('resources', validate=False) key= bucket.get_key('filename.txt') cardinal.set_contents_from_string("SAMPLE TEXT") key.set_canned_acl('public-read') ~ Answered on 2018-eleven-xiii 19:59:12
Source: https://syntaxfix.com/question/19151/how-to-upload-a-file-to-directory-in-s3-bucket-using-boto
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