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Make s3upload function return the UUID

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Generating the UUID can be can be contained within this function,
thus any other part of the code using it doesn’t have to do the
ID-generating stuff itself.
This commit is contained in:
Chris Hill-Scott
2016-05-15 07:47:50 +01:00
parent b35f2d7e61
commit e84436d0e1
4 changed files with 25 additions and 16 deletions
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6
app/main/uploader.py
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@@ -1,3 +1,4 @@
import uuid
import botocore
from boto3 import resource
from flask import current_app
@@ -5,7 +6,7 @@ from flask import current_app
FILE_LOCATION_STRUCTURE = 'service-{}-notify/{}.csv'
def s3upload(upload_id, service_id, filedata, region):
def s3upload(service_id, filedata, region):
s3 = resource('s3')
bucket_name = current_app.config['CSV_UPLOAD_BUCKET_NAME']
contents = filedata['data']
@@ -27,10 +28,13 @@ def s3upload(upload_id, service_id, filedata, region):
s3.create_bucket(Bucket=bucket_name,
CreateBucketConfiguration={'LocationConstraint': region})
upload_id = str(uuid.uuid4())
upload_file_name = FILE_LOCATION_STRUCTURE.format(service_id, upload_id)
key = s3.Object(bucket_name, upload_file_name)
key.put(Body=contents, ServerSideEncryption='AES256')
return upload_id
def s3download(service_id, upload_id):
contents = ''