How to get AWS S3 object Location/URL using python 3.8?

Question

I am uploading a file to AWS S3 using AWS Lambda function (Python3.8) with the following code.

   file_obj = open(filename, 'rb')

   s3_upload = s3.put_object( Bucket="aaa", Key="aaa.png", Body=file_obj)
   
   return {
   'statusCode': 200,
   'body': json.dumps("Executed Successfully")
   }

I want to get the location/url of the S3 object in return. In Node.js we use the .location parameter for getting the object location/url.

Any idea how to do this using python 3.8?

Marcin · Accepted Answer · 2020-11-24 05:40:30Z

3

The url of S3 objects has known format and follows Virtual hosted style access:

https://bucket-name.s3.Region.amazonaws.com/keyname

Thus, you can construct the url yourself:

bucket_name = 'aaa'
aws_region = boto3.session.Session().region_name
object_key = 'aaa.png'

s3_url = f"https://{bucket_name}.s3.{aws_region}.amazonaws.com/{object_key}"

return {
   'statusCode': 200,
   'body': json.dumps({'s3_url': s3_url})
}

answered Nov 24, 2020 at 5:40

Marcin

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2 Comments

Muhammad Arsalan Hassan Over a year ago

It's returning null

Muhammad Arsalan Hassan Over a year ago

It was some issue on my end. It's working fine now. Thanks for your help.

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