Fix lambda handler name

Grant Cooksey · Grant Cooksey · commit 2fc184071384 · 2019-01-26T10:23:55.000-05:00
diff --git a/README.md b/README.md
@@ -162,7 +162,7 @@ want to configure the function further, such as adding environment variables or
 Alternatively to doing this in your cli, you can create the function inside the aws lambda console.
 
 ```bash
-aws lambda create-function --function-name FUNCTION_NAME --role IAM_ROLE_ARN --handler app.lambda_handler --runtime python3.6 --code S3Bucket=S3_BUCKET_NAME,S3Key=FUNCTION_NAME.zip
+aws lambda create-function --function-name FUNCTION_NAME --role IAM_ROLE_ARN --handler service.app.lambda_handler --runtime python3.6 --code S3Bucket=S3_BUCKET_NAME,S3Key=FUNCTION_NAME.zip
 ```
 
 The function has been created. You can [invoke](https://docs.aws.amazon.com/cli/latest/reference/lambda/invoke.html)
diff --git a/examples/list_objects_s3/src/list_objects_s3/app.py b/examples/list_objects_s3/src/list_objects_s3/app.py
@@ -14,6 +14,7 @@ def lambda_handler(event, context):
         logger.info(f'Calling out to {S3_BUCKET} bucket to list objects')
         s3_client = boto3.client('s3')
 
+        # Get all the objects from the bucket. Max 1000
         objects_in_bucket = s3_client.list_objects(Bucket=S3_BUCKET)
 
         logger.info('Found {} objects in the bucket. Printing a sample...'.format(len(objects_in_bucket['Contents'])))
