create utils.py in AWS lambda

Question

I had a def hello() function in my home/file.py file. I created a home/common/utils.pyfile and moved the function there. Now, I want to import it in my file file.py.

I imported it like this: from utils import hello and from common.utils import hello and the import in my file doesn't throw an error. However, when I run it on AWS Lambda, I get an error that:

Runtime.ImportModuleError: Unable to import module 'file': No module named 'utils'

How can I fix this? without having to use Ec2 or something...

data "archive_file" "file_zip" {
  type             = "zip"
  source_file      = "${path.module}/src/file.py"
  output_file_mode = "0666"
  output_path      = "${path.module}/bin/file.zip"
}

What is the hierarchy of files in the packaged ZIP file that you uploaded to Lambda? — jarmod
– jarmod, Commented Nov 19, 2021 at 15:39
Not sure what you mean..I just run terraform init, apply and it happens automatically. How can I check the structure @jarmod — x89
– x89, Commented Nov 19, 2021 at 15:41
When you deploy a Lambda function, regardless of which IaC tool you use, you provide a packaged deployment artifact (a ZIP file of your code and dependencies). You typically do that by zipping your code and dependencies into lambda.zip, for example, then indicating that file in the aws_lambda_function resource in your Terraform template, then Terraform takes it and deploys it. — jarmod
– jarmod, Commented Nov 19, 2021 at 15:47
I updated the qs with how I zip the file content. not sure if this is what u meant by the hierarchy @jarmod — x89
– x89, Commented Nov 19, 2021 at 15:52

jarmod · Accepted Answer · 2021-11-20 00:05:44Z

The deployment package that you're uploading only contains your main Python script (file.py). Specifically, it does not include any dependencies such as common/utils.py. That's why the import fails when the code runs in Lambda.

Modify the creation of your deployment package (file.zip) so that it includes all needed dependencies.

For example:

data "archive_file" "file_zip" {
  type             = "zip"
  output_file_mode = "0666"
  output_path      = "${path.module}/bin/file.zip"

  source {
    content  = file("${path.module}/src/file.py")
    filename = "file.py"
  }

  source {
    content  = file("${path.module}/src/common/utils.py")
    filename = "common/utils.py"
  }
}

If all of your files happen to be in a single folder then you can use source_dir instead of indicating the individual files.

Note: I don't use Terraform so the file(...) with embedded interpolation may not be 100% correct, but you get the idea.

Parth · Accepted Answer · 2022-07-05 08:29:07Z

First of all, properly follow this standard URL:- https://docs.aws.amazon.com/lambda/latest/dg/python-package.html (refer section with title:- Deployment package with dependencies)

Now, if you notice, in the end of section,

zip -g my-deployment-package.zip lambda_function.py

Follow the same command for your utils file:-

zip -g my-deployment-package.zip common/    
zip -g my-deployment-package.zip common/utils.py

Ensure that, in lambda_function, you are using proper import statement like:-

from common.utils import util_function_name

Now, you can upload this zip and test for yourself. It should run.

Hope this helps.

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