PYTHON

upload data to s3 bucket python

import boto
import boto.s3
import sys
from boto.s3.key import Key

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"
print 'Uploading %s to Amazon S3 bucket %s' % 
   (testfile, bucket_name)

def percent_cb(complete, total):
    sys.stdout.write('.')
    sys.stdout.flush()


k = Key(bucket)
k.key = 'my test file'
k.set_contents_from_filename(testfile,
    cb=percent_cb, num_cb=10)

upload file to s3 python

# easy way
import boto3

s3 = boto3.resource(
    service_name='s3',
    region_name='Your region',
    aws_access_key_id='Your id',
    aws_secret_access_key="Your Key")

bucket = s3.Bucket('your bucket name')

bucket.upload_file(Filename='temp.csv', Key='temp.csv')
# file path __________↑                 ↑ 
# (which you want to upload)            |
#                                       |______path (where you want to upload inside bucket)

python upload file to s3

import boto3

s3 = boto3.resource('s3')
BUCKET = "test"

s3.Bucket(BUCKET).upload_file("your/local/file", "dump/file")

upload image to s3 python

The AWS SDK for Python provides a pair of methods to upload a file to an S3 bucket. The upload_file method accepts a file name, a bucket name, and an object name.