Develop Reference

How do I directly save images from a url to my aws bucket (python / boto3)? [duplicate]

I'm working in a Python web environment and I can simply upload a file from the filesystem to S3 using boto's key.set_contents_from_filename(path/to/file). However, I'd like to upload an image that is already on the web (say https://pbs.twimg.com/media/A9h_htACIAAaCf6.jpg:large).
Should I somehow download the image to the filesystem, and then upload it to S3 using boto as usual, then delete the image?
What would be ideal is if there is a way to get boto's key.set_contents_from_file or some other command that would accept a URL and nicely stream the image to S3 without having to explicitly download a file copy to my server.
def upload(url):
try:
conn = boto.connect_s3(settings.AWS_ACCESS_KEY_ID, settings.AWS_SECRET_ACCESS_KEY)
bucket_name = settings.AWS_STORAGE_BUCKET_NAME
bucket = conn.get_bucket(bucket_name)
k = Key(bucket)
k.key = "test"
k.set_contents_from_file(url)
k.make_public()
return "Success?"
except Exception, e:
return e
Using set_contents_from_file, as above, I get a "string object has no attribute 'tell'" error. Using set_contents_from_filename with the url, I get a No such file or directory error . The boto storage documentation leaves off at uploading local files and does not mention uploading files stored remotely.

Here is how I did it with requests, the key being to set stream=True when initially making the request, and uploading to s3 using the upload.fileobj() method:
import requests
import boto3
url = "https://upload.wikimedia.org/wikipedia/en/a/a9/Example.jpg"
r = requests.get(url, stream=True)
session = boto3.Session()
s3 = session.resource('s3')
bucket_name = 'your-bucket-name'
key = 'your-key-name' # key is the name of file on your bucket
bucket = s3.Bucket(bucket_name)
bucket.upload_fileobj(r.raw, key)

Ok, from #garnaat, it doesn't sound like S3 currently allows uploads by url. I managed to upload remote images to S3 by reading them into memory only. This works.
def upload(url):
try:
conn = boto.connect_s3(settings.AWS_ACCESS_KEY_ID, settings.AWS_SECRET_ACCESS_KEY)
bucket_name = settings.AWS_STORAGE_BUCKET_NAME
bucket = conn.get_bucket(bucket_name)
k = Key(bucket)
k.key = url.split('/')[::-1][0] # In my situation, ids at the end are unique
file_object = urllib2.urlopen(url) # 'Like' a file object
fp = StringIO.StringIO(file_object.read()) # Wrap object
k.set_contents_from_file(fp)
return "Success"
except Exception, e:
return e
Also thanks to How can I create a GzipFile instance from the “file-like object” that urllib.urlopen() returns?

For a 2017-relevant answer to this question which uses the official 'boto3' package (instead of the old 'boto' package from the original answer):
Python 3.5
If you're on a clean Python install, pip install both packages first:
pip install boto3
pip install requests
import boto3
import requests
# Uses the creds in ~/.aws/credentials
s3 = boto3.resource('s3')
bucket_name_to_upload_image_to = 'photos'
s3_image_filename = 'test_s3_image.png'
internet_image_url = 'https://docs.python.org/3.7/_static/py.png'
# Do this as a quick and easy check to make sure your S3 access is OK
for bucket in s3.buckets.all():
if bucket.name == bucket_name_to_upload_image_to:
print('Good to go. Found the bucket to upload the image into.')
good_to_go = True
if not good_to_go:
print('Not seeing your s3 bucket, might want to double check permissions in IAM')
# Given an Internet-accessible URL, download the image and upload it to S3,
# without needing to persist the image to disk locally
req_for_image = requests.get(internet_image_url, stream=True)
file_object_from_req = req_for_image.raw
req_data = file_object_from_req.read()
# Do the actual upload to s3
s3.Bucket(bucket_name_to_upload_image_to).put_object(Key=s3_image_filename, Body=req_data)

Unfortunately, there really isn't any way to do this. At least not at the moment. We could add a method to boto, say set_contents_from_url, but that method would still have to download the file to the local machine and then upload it. It might still be a convenient method but it wouldn't save you anything.
In order to do what you really want to do, we would need to have some capability on the S3 service itself that would allow us to pass it the URL and have it store the URL to a bucket for us. That sounds like a pretty useful feature. You might want to post that to the S3 forums.

A simple 3-lines implementation that works on a lambda out-of-the-box:
import boto3
import requests
s3_object = boto3.resource('s3').Object(bucket_name, object_key)
with requests.get(url, stream=True) as r:
s3_object.put(Body=r.content)
The source for the .get part comes straight from the requests documentation

from io import BytesIO
def send_image_to_s3(url, name):
print("sending image")
bucket_name = 'XXX'
AWS_SECRET_ACCESS_KEY = "XXX"
AWS_ACCESS_KEY_ID = "XXX"
s3 = boto3.client('s3', aws_access_key_id=AWS_ACCESS_KEY_ID,
aws_secret_access_key=AWS_SECRET_ACCESS_KEY)
response = requests.get(url)
img = BytesIO(response.content)
file_name = f'path/{name}'
print('sending {}'.format(file_name))
r = s3.upload_fileobj(img, bucket_name, file_name)
s3_path = 'path/' + name
return s3_path

I have tried as following with boto3 and it works me:
import boto3;
import contextlib;
import requests;
from io import BytesIO;
s3 = boto3.resource('s3');
s3Client = boto3.client('s3')
for bucket in s3.buckets.all():
print(bucket.name)
url = "#resource url";
with contextlib.closing(requests.get(url, stream=True, verify=False)) as response:
# Set up file stream from response content.
fp = BytesIO(response.content)
# Upload data to S3
s3Client.upload_fileobj(fp, 'aws-books', 'reviews_Electronics_5.json.gz')

Using the boto3 upload_fileobj method, you can stream a file to an S3 bucket, without saving to disk. Here is my function:
import boto3
import StringIO
import contextlib
import requests
def upload(url):
# Get the service client
s3 = boto3.client('s3')
# Rember to se stream = True.
with contextlib.closing(requests.get(url, stream=True, verify=False)) as response:
# Set up file stream from response content.
fp = StringIO.StringIO(response.content)
# Upload data to S3
s3.upload_fileobj(fp, 'my-bucket', 'my-dir/' + url.split('/')[-1])

S3 doesn't support remote upload as of now it seems. You may use the below class for uploading an image to S3. The upload method here first tries to download the image and keeps it in memory for sometime until it gets uploaded. To be able to connect to S3 you will have to install AWS CLI using command pip install awscli, then enter few credentials using command aws configure:
import urllib3
import uuid
from pathlib import Path
from io import BytesIO
from errors import custom_exceptions as cex
BUCKET_NAME = "xxx.yyy.zzz"
POSTERS_BASE_PATH = "assets/wallcontent"
CLOUDFRONT_BASE_URL = "https://xxx.cloudfront.net/"
class S3(object):
def __init__(self):
self.client = boto3.client('s3')
self.bucket_name = BUCKET_NAME
self.posters_base_path = POSTERS_BASE_PATH
def __download_image(self, url):
manager = urllib3.PoolManager()
try:
res = manager.request('GET', url)
except Exception:
print("Could not download the image from URL: ", url)
raise cex.ImageDownloadFailed
return BytesIO(res.data) # any file-like object that implements read()
def upload_image(self, url):
try:
image_file = self.__download_image(url)
except cex.ImageDownloadFailed:
raise cex.ImageUploadFailed
extension = Path(url).suffix
id = uuid.uuid1().hex + extension
final_path = self.posters_base_path + "/" + id
try:
self.client.upload_fileobj(image_file,
self.bucket_name,
final_path
)
except Exception:
print("Image Upload Error for URL: ", url)
raise cex.ImageUploadFailed
return CLOUDFRONT_BASE_URL + id

import boto
from boto.s3.key import Key
from boto.s3.connection import OrdinaryCallingFormat
from urllib import urlopen
def upload_images_s3(img_url):
try:
connection = boto.connect_s3('access_key', 'secret_key', calling_format=OrdinaryCallingFormat())
bucket = connection.get_bucket('boto-demo-1519388451')
file_obj = Key(bucket)
file_obj.key = img_url.split('/')[::-1][0]
fp = urlopen(img_url)
result = file_obj.set_contents_from_string(fp.read())
except Exception, e:
return e

Upload multiple files from S3 to Frame IO

On file upload in S3, I am triggering lambda function which will generate s3 url and create file in Frame IO. Whenever I am trying to upload many files at once in S3, file is not creating properly in Frame IO and throwing Preview Unsupported Error (for mp4 files which is supported by default). To fix this issue, I tried to use index as a request parameter which worked out only on 2 or 3 files upload. If I am trying to upload more files, the same error arise. Please find the lambda function code below
import requests
import boto3
import json
import urllib.parse
import mimetypes
from botocore.config import Config
import os
s3_client = boto3.client('s3', config = Config(signature_version='s3v4'))
client = boto3.client('ssm')
def lambda_handler(event, context):
print(event)
bucket = event['Records'][0]['s3']['bucket']['name']
key = urllib.parse.unquote_plus(event['Records'][0]['s3']['object']['key'], encoding='utf-8')
if not key.endswith('/'):
if key.find('/') >= 0:
temp_key = key.rsplit('/', 1)
key = temp_key[1]
print(key)
size = event['Records'][0]['s3']['object']['size']
frameioIndex = int(client.get_parameter(Name='/frameio/asset/index_Dev')['Parameter']['Value']) - 1
print(frameioIndex)
s3_url = s3_client.generate_presigned_url("get_object", Params={"Bucket": bucket, "Key": key})
response = requests.post(os.environ['FRAMEIO_BASE_API_URL'] + "assets" + "/" + os.environ['FRAMEIO_PROJECT_ID'] + "/" + "children",data=json.dumps({"type": "file","name": key,"filesize": size,"filetype": mimetypes.guess_type(key)[0],"source": {"url": s3_url},"index": frameioIndex}), headers={"Authorization":"Bearer " + os.environ['FRAMEIO_TOKEN'], "Content-type": "application/json"}) client.put_parameter(Name='/frameio/asset/index_Dev',Value=str(frameioIndex),Type='String',Overwrite=True)
print(response)
return {
'statusCode': 200,
'body': json.dumps('Successfully uploaded the asset!')
}
return {
'statusCode': 200,
'body': json.dumps('Uploaded object is not a file!')
}

The issue resolved by changing variable name 'key' to 'filename' in line number 14 (key = temp_key[1]) and used filename in requests API. The above issue occurred as I tried to override the filename and passing it to generate_presigned_url method to generate s3 url.

AWS presigned URLS location constraint is incompatible for the region specific endpoint this request was sent to

I am using a lambda to create a pre-signed URL to download files that land in an S3 bucket -
the code works and I get a URL but when trying to access it I get
af-south-1 location constraint is incompatible for the region-specific endpoint this request was sent to.
both the bucket and the lambda are in the same region
I'm at a loss as to what is actually happening any ideas or solutions would be greatly appreciated.
my code is below
import json
import boto3
import boto3.session
def lambda_handler(event, context):
session = boto3.session.Session(region_name='af-south-1')
s3 = session.client('s3')
for record in event['Records']:
bucket = record['s3']['bucket']['name']
key = record['s3']['object']['key']
url = s3.generate_presigned_url(ClientMethod='get_object',
Params={'Bucket': bucket,
'Key': key}, ExpiresIn = 400)
print (url)```

Set an endpoint_url=https://s3.af-south-1.amazonaws.com while generating the s3_client
s3_client = session.client('s3',
region_name='af-south-1',
endpoint_url='https://s3.af-south-1.amazonaws.com')

Could you please try using the boto3 client directly rather than via the session, and generate the pre-signed url :
import boto3
import requests
# Get the service client.
s3 = boto3.client('s3',region_name='af-south-1')
# Generate the URL to get 'key-name' from 'bucket-name'
url = s3.generate_presigned_url(
ClientMethod='get_object',
Params={
'Bucket': 'bucket-name',
'Key': 'key-name'
}
)
You could also have a look at these 1 & 2, which resembles the same issue.

How to return byte array from AWS Lambda API gateway?

I am a beginner so I am hoping to get some help here.
I want create a lambda function (written in Python) that is able to read an image stored in S3 then return the image as a binary file (eg. a byte array). The lambda function is triggered by an API gateway.
Right now, I have setup the API gateway to trigger the Lambda function and it can return a hello message. I also have a gif image stored in a S3 bucket.
import base64
import json
import boto3
s3 = boto.client('s3')
def lambda_handler(event, context):
# TODO implement
bucket = 'mybucket'
key = 'myimage.gif'
s3.get_object(Bucket=bucket, Key=key)['Body'].read()
return {
"statusCode": 200,
"body": json.dumps('Hello from AWS Lambda!!')
}
I really have no idea how to continue. Can anyone advise? Thanks in advance!

you can return Base64 encoded data from your Lambda function with appropriate headers.
Here the updated Lambda function:
import base64
import boto3
s3 = boto3.client('s3')
def lambda_handler(event, context):
bucket = 'mybucket'
key = 'myimage.gif'
image_bytes = s3.get_object(Bucket=bucket, Key=key)['Body'].read()
# We will now convert this image to Base64 string
image_base64 = base64.b64encode(image_bytes)
return {'statusCode': 200,
# Providing API Gateway the headers for the response
'headers': {'Content-Type': 'image/gif'},
# The image in a Base64 encoded string
'body': image_base64,
'isBase64Encoded': True}
For further details and step by step guide, you can refer to this official blog.

Upload image available at public URL to S3 using boto

I'm working in a Python web environment and I can simply upload a file from the filesystem to S3 using boto's key.set_contents_from_filename(path/to/file). However, I'd like to upload an image that is already on the web (say https://pbs.twimg.com/media/A9h_htACIAAaCf6.jpg:large).
Should I somehow download the image to the filesystem, and then upload it to S3 using boto as usual, then delete the image?
What would be ideal is if there is a way to get boto's key.set_contents_from_file or some other command that would accept a URL and nicely stream the image to S3 without having to explicitly download a file copy to my server.
def upload(url):
try:
conn = boto.connect_s3(settings.AWS_ACCESS_KEY_ID, settings.AWS_SECRET_ACCESS_KEY)
bucket_name = settings.AWS_STORAGE_BUCKET_NAME
bucket = conn.get_bucket(bucket_name)
k = Key(bucket)
k.key = "test"
k.set_contents_from_file(url)
k.make_public()
return "Success?"
except Exception, e:
return e
Using set_contents_from_file, as above, I get a "string object has no attribute 'tell'" error. Using set_contents_from_filename with the url, I get a No such file or directory error . The boto storage documentation leaves off at uploading local files and does not mention uploading files stored remotely.

Here is how I did it with requests, the key being to set stream=True when initially making the request, and uploading to s3 using the upload.fileobj() method:
import requests
import boto3
url = "https://upload.wikimedia.org/wikipedia/en/a/a9/Example.jpg"
r = requests.get(url, stream=True)
session = boto3.Session()
s3 = session.resource('s3')
bucket_name = 'your-bucket-name'
key = 'your-key-name' # key is the name of file on your bucket
bucket = s3.Bucket(bucket_name)
bucket.upload_fileobj(r.raw, key)

Ok, from #garnaat, it doesn't sound like S3 currently allows uploads by url. I managed to upload remote images to S3 by reading them into memory only. This works.
def upload(url):
try:
conn = boto.connect_s3(settings.AWS_ACCESS_KEY_ID, settings.AWS_SECRET_ACCESS_KEY)
bucket_name = settings.AWS_STORAGE_BUCKET_NAME
bucket = conn.get_bucket(bucket_name)
k = Key(bucket)
k.key = url.split('/')[::-1][0] # In my situation, ids at the end are unique
file_object = urllib2.urlopen(url) # 'Like' a file object
fp = StringIO.StringIO(file_object.read()) # Wrap object
k.set_contents_from_file(fp)
return "Success"
except Exception, e:
return e
Also thanks to How can I create a GzipFile instance from the “file-like object” that urllib.urlopen() returns?

For a 2017-relevant answer to this question which uses the official 'boto3' package (instead of the old 'boto' package from the original answer):
Python 3.5
If you're on a clean Python install, pip install both packages first:
pip install boto3
pip install requests
import boto3
import requests
# Uses the creds in ~/.aws/credentials
s3 = boto3.resource('s3')
bucket_name_to_upload_image_to = 'photos'
s3_image_filename = 'test_s3_image.png'
internet_image_url = 'https://docs.python.org/3.7/_static/py.png'
# Do this as a quick and easy check to make sure your S3 access is OK
for bucket in s3.buckets.all():
if bucket.name == bucket_name_to_upload_image_to:
print('Good to go. Found the bucket to upload the image into.')
good_to_go = True
if not good_to_go:
print('Not seeing your s3 bucket, might want to double check permissions in IAM')
# Given an Internet-accessible URL, download the image and upload it to S3,
# without needing to persist the image to disk locally
req_for_image = requests.get(internet_image_url, stream=True)
file_object_from_req = req_for_image.raw
req_data = file_object_from_req.read()
# Do the actual upload to s3
s3.Bucket(bucket_name_to_upload_image_to).put_object(Key=s3_image_filename, Body=req_data)

Unfortunately, there really isn't any way to do this. At least not at the moment. We could add a method to boto, say set_contents_from_url, but that method would still have to download the file to the local machine and then upload it. It might still be a convenient method but it wouldn't save you anything.
In order to do what you really want to do, we would need to have some capability on the S3 service itself that would allow us to pass it the URL and have it store the URL to a bucket for us. That sounds like a pretty useful feature. You might want to post that to the S3 forums.

A simple 3-lines implementation that works on a lambda out-of-the-box:
import boto3
import requests
s3_object = boto3.resource('s3').Object(bucket_name, object_key)
with requests.get(url, stream=True) as r:
s3_object.put(Body=r.content)
The source for the .get part comes straight from the requests documentation

from io import BytesIO
def send_image_to_s3(url, name):
print("sending image")
bucket_name = 'XXX'
AWS_SECRET_ACCESS_KEY = "XXX"
AWS_ACCESS_KEY_ID = "XXX"
s3 = boto3.client('s3', aws_access_key_id=AWS_ACCESS_KEY_ID,
aws_secret_access_key=AWS_SECRET_ACCESS_KEY)
response = requests.get(url)
img = BytesIO(response.content)
file_name = f'path/{name}'
print('sending {}'.format(file_name))
r = s3.upload_fileobj(img, bucket_name, file_name)
s3_path = 'path/' + name
return s3_path

I have tried as following with boto3 and it works me:
import boto3;
import contextlib;
import requests;
from io import BytesIO;
s3 = boto3.resource('s3');
s3Client = boto3.client('s3')
for bucket in s3.buckets.all():
print(bucket.name)
url = "#resource url";
with contextlib.closing(requests.get(url, stream=True, verify=False)) as response:
# Set up file stream from response content.
fp = BytesIO(response.content)
# Upload data to S3
s3Client.upload_fileobj(fp, 'aws-books', 'reviews_Electronics_5.json.gz')

Using the boto3 upload_fileobj method, you can stream a file to an S3 bucket, without saving to disk. Here is my function:
import boto3
import StringIO
import contextlib
import requests
def upload(url):
# Get the service client
s3 = boto3.client('s3')
# Rember to se stream = True.
with contextlib.closing(requests.get(url, stream=True, verify=False)) as response:
# Set up file stream from response content.
fp = StringIO.StringIO(response.content)
# Upload data to S3
s3.upload_fileobj(fp, 'my-bucket', 'my-dir/' + url.split('/')[-1])

S3 doesn't support remote upload as of now it seems. You may use the below class for uploading an image to S3. The upload method here first tries to download the image and keeps it in memory for sometime until it gets uploaded. To be able to connect to S3 you will have to install AWS CLI using command pip install awscli, then enter few credentials using command aws configure:
import urllib3
import uuid
from pathlib import Path
from io import BytesIO
from errors import custom_exceptions as cex
BUCKET_NAME = "xxx.yyy.zzz"
POSTERS_BASE_PATH = "assets/wallcontent"
CLOUDFRONT_BASE_URL = "https://xxx.cloudfront.net/"
class S3(object):
def __init__(self):
self.client = boto3.client('s3')
self.bucket_name = BUCKET_NAME
self.posters_base_path = POSTERS_BASE_PATH
def __download_image(self, url):
manager = urllib3.PoolManager()
try:
res = manager.request('GET', url)
except Exception:
print("Could not download the image from URL: ", url)
raise cex.ImageDownloadFailed
return BytesIO(res.data) # any file-like object that implements read()
def upload_image(self, url):
try:
image_file = self.__download_image(url)
except cex.ImageDownloadFailed:
raise cex.ImageUploadFailed
extension = Path(url).suffix
id = uuid.uuid1().hex + extension
final_path = self.posters_base_path + "/" + id
try:
self.client.upload_fileobj(image_file,
self.bucket_name,
final_path
)
except Exception:
print("Image Upload Error for URL: ", url)
raise cex.ImageUploadFailed
return CLOUDFRONT_BASE_URL + id

import boto
from boto.s3.key import Key
from boto.s3.connection import OrdinaryCallingFormat
from urllib import urlopen
def upload_images_s3(img_url):
try:
connection = boto.connect_s3('access_key', 'secret_key', calling_format=OrdinaryCallingFormat())
bucket = connection.get_bucket('boto-demo-1519388451')
file_obj = Key(bucket)
file_obj.key = img_url.split('/')[::-1][0]
fp = urlopen(img_url)
result = file_obj.set_contents_from_string(fp.read())
except Exception, e:
return e