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Fix for handling S3 deployment arguments #94

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javamonkey79 merged 1 commit into from
Feb 15, 2018
Merged

Fix for handling S3 deployment arguments #94

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fix for handling S3 lambda deployment arguments
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@frnsys
frnsys committed Feb 7, 2018
commit 419ea8785e17f2fc87a065faae48cffc572e04f9
8 changes: 4 additions & 4 deletions aws_lambda/aws_lambda.py
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Original file line number Diff line number Diff line change
Expand Up @@ -139,9 +139,9 @@ def deploy_s3(
use_s3 = True
s3_file = upload_s3(cfg, path_to_zip_file, use_s3)
if function_exists(cfg, cfg.get('function_name')):
update_function(cfg, path_to_zip_file, use_s3, s3_file)
update_function(cfg, path_to_zip_file, use_s3=use_s3, s3_file=s3_file)
else:
create_function(cfg, path_to_zip_file, use_s3, s3_file)
create_function(cfg, path_to_zip_file, use_s3=use_s3, s3_file=s3_file)


def upload(
Expand Down Expand Up @@ -459,7 +459,7 @@ def get_client(client, aws_access_key_id, aws_secret_access_key, region=None):
)


def create_function(cfg, path_to_zip_file, *use_s3, **s3_file):
def create_function(cfg, path_to_zip_file, use_s3=False, s3_file=None):
"""Register and upload a function to AWS Lambda."""

print('Creating your new Lambda function')
Expand Down Expand Up @@ -531,7 +531,7 @@ def create_function(cfg, path_to_zip_file, *use_s3, **s3_file):
client.create_function(**kwargs)


def update_function(cfg, path_to_zip_file, *use_s3, **s3_file):
def update_function(cfg, path_to_zip_file, use_s3=False, s3_file=None):
"""Updates the code of an existing Lambda function"""

print('Updating your Lambda function')
Expand Down