How can I create an AWS Lambda function using the AWS CLI?

Question

I am trying to create an AWS Lambda function using the command

aws lambda create-function \
  --function-name foo\
  --runtime nodejs\
  --role lambda_basic_execution \
  --handler asdf --zip-file "fileb:://boom.zip"

I have a file called boom.zip available in the directory. But I cannot deploy using the above command.

The failure message I get is

--zip-file must be a file with the fileb:// prefix.

Does anyone have a working example to create a lambda function using the AWS CLI?

In your snippet, you have "fileb:://" which is incorrect based on the error message. — Matt Houser
I haven't got this working either. The example I'm working from has a longer --role arn:aws:iam::$account_id:role/service-role/lambda_basic_execution per aws.amazon.com/blogs/compute/… — MarkHu
double check your zip file. I just ran into this which got me to this thread, and come to find out, my build created an empty zip due to an error — Scott

helloV helloV · Accepted Answer · 2015-12-18T21:43:57

You have an extra colon ':' in the file spec.

$ aws lambda create-function --function-name foo --runtime nodejs --role lambda_basic_execution --handler asdf --zip-file "fileb:://boom.zip"

--zip-file must be a file with the fileb:// prefix.
Example usage:  --zip-file fileb://path/to/file.zip

$ aws lambda create-function --function-name foo --runtime nodejs --role lambda_basic_execution --handler asdf --zip-file "fileb://boom.zip"

Error parsing parameter '--zip-file': Unable to load paramfile fileb://boom.zip: [Errno 2] No such file or directory: 'boom.zip'

How can I create an AWS Lambda function using the AWS CLI?

3 Answers