Deploy lambda function inline

Question

I have a python function that I can deploy through S3 bucket. But it is possible to deploy a function "inline"...

https://docs.aws.amazon.com/AWSCloudFormation/latest/UserGuide/aws-properties-lambda-function-code.html

But I do not see any clear instructions about how to do this. I do not want to use S3 bucket.

jens walter · Accepted Answer · 2019-07-14 14:38:43Z

12

You can deploy an AWS Lambda function inline within a CloudFormation template through the following YAML syntax.

  LambdaFunction:
    Type: AWS::Lambda::Function
    Properties:
      Code:
        ZipFile: >
          def handler(event, context):
            print('hello from lambda')
      Handler: index.handler
      Role:
        Fn::GetAtt: [ LambdaRole , "Arn" ]
      Runtime: python3.6
      Timeout: 60

answered Jul 14, 2019 at 14:38

jens walter

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4 Comments

shantanuo Over a year ago

Just for the sake of reference, Can you point to the documentation of Code: ZipFile > section?

jens walter Over a year ago

The documentation you already linked states, that the Zipfile property allows for inline code. The rest is basically YAML syntax (and a lot of trial and error).

lapkritinis Over a year ago

@shantanuo Here is AWS example with inline code docs.aws.amazon.com/AWSCloudFormation/latest/UserGuide/…

Raheel Over a year ago

I am wondering why you have used > insead of a pipe | ?

Divanshu Aggarwal · Accepted Answer · 2019-07-14 13:42:08Z

-1

When you create a lambda function you will see option to choose language.
Choose python
Create function
As soon as function is created you will see edit inline option and a file of your lambda_function_name/lambda_function.py with code
```
import json.
def lambda_handler(event, context):
return {
 'statusCode': 200,
 'body': json.dumps('Hello from Lambda!')
}
```

you can edit this based upon requirement.This is inline.

answered Jul 14, 2019 at 13:42

Divanshu Aggarwal

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