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ldipotetjob/makefile_for_update_deploy_lambda_function

Last active May 22, 2025 12:38
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Update and Deploy a lambda function to AWS Lambda
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makefile_for_update_deploy_lambda_function
ROLE_ARN=$(ROLE)
ZIP_FILE=deployment-package.zip
REGION=eu-west-1
# run under this project structure: https://github.com/ldipotetjob/AWS/blob/main/serverless/lambda/project/structure.md
# keep in mind your security tokens export in your terminal:
# export AWS_ACCESS_KEY_ID=${token_access_key}
# export AWS_SECRET_ACCESS_KEY=${token_secret_key}
# @see: https://makefiletutorial.com/
# example of use: make all HANDLER=my_lambda_function ROLE=execution_role_of_the_lambda_function
FUNCTION_NAME=$(HANDLER)
all: help build deploy clean
help:
echo "deploying " $(HANDLER)
build:
mkdir -p deployment-package
cp -r src/$(HANDLER)/. src/standard_utils.py deployment-package/
if [ -f requirements.txt ]; then pip install -r requirements.txt --target deployment-package/; fi
cd deployment-package && zip -r9 ../$(ZIP_FILE) .
deploy:
echo "Deploying lambda function:" $(HANDLER)
aws lambda update-function-code --function-name $(FUNCTION_NAME) --zip-file fileb://$(ZIP_FILE) --region $(REGION)
clean:
rm -rf deployment-package $(ZIP_FILE)
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