Upload to Amazon S3 using Boto3 and return public url

Question

Iam trying to upload files to s3 using Boto3 and make that uploaded file public and return it as a url.

class UtilResource(BaseZMPResource): class Meta(BaseZMPResource.Meta):     queryset = Configuration.objects.none()     resource_name = 'util_resource'     allowed_methods = ['get']  def post_list(self, request, **kwargs):      fileToUpload = request.FILES     # write code to upload to amazone s3     # see: https://boto3.readthedocs.org/en/latest/reference/services/s3.html       self.session = Session(aws_access_key_id=settings.AWS_KEY_ID,                   aws_secret_access_key=settings.AWS_ACCESS_KEY,                   region_name=settings.AWS_REGION)      client = self.session.client('s3')     client.upload_file('zango-static','fileToUpload')       url = "some/test/url"     return self.create_response(request, {         'url': url // return's public url of uploaded file      }) 

I searched whole documentation I couldn't find any links which describes how to do this can someone explain or provide any resource where I can find the soultion?

Answer 1

I'm in the same situation. Not able to find anything in the Boto3 docs beyond generate_presigned_url which is not what I need in my case since I have public readable S3 Objects.

The best I came up with is:

bucket_location = boto3.client('s3').get_bucket_location(Bucket=s3_bucket_name) object_url = "https://s3-{0}.amazonaws.com/{1}/{2}".format(     bucket_location['LocationConstraint'],     s3_bucket_name,     key_name) 

You might try posting on the boto3 github issues list for a better solution.