Can we create lambda function using grunt instead of amazon console? - aws-lambda

Question

I am creating a zip file for deployment to lambda using https://github.com/Tim-B/grunt-aws-lambda but when deploying to aws lambda I need to create the function first in amazon console. Can we create a function using grunt instead of amazon console? Thank you.

Answer 1

You can create the function from grunt using AWS JavaScript SDK for Lambda.
Use the createFunction method as shown below.
/* This example creates a Lambda function. */
var params = {
Code: {
},
Description: "",
FunctionName: "MyFunction",
Handler: "souce_file.handler_name", // is of the form of the name of your source file and then name of your function handler
MemorySize: 128,
Publish: true,
Role: "arn:aws:iam::123456789012:role/service-role/role-name", // replace with the actual arn of the execution role you created
Runtime: "nodejs4.3",
Timeout: 15,
VpcConfig: {
}
};
lambda.createFunction(params, function(err, data) {
if (err) console.log(err, err.stack); // an error occurred
else console.log(data); // successful response
/*
data = {
CodeSha256: "",
CodeSize: 123,
Description: "",
FunctionArn: "arn:aws:lambda:us-west-2:123456789012:function:MyFunction",
FunctionName: "MyFunction",
Handler: "source_file.handler_name",
LastModified: "2016-11-21T19:49:20.006+0000",
MemorySize: 128,
Role: "arn:aws:iam::123456789012:role/service-role/role-name",
Runtime: "nodejs4.3",
Timeout: 123,
Version: "1",
VpcConfig: {
}
}
*/
});
Note: You can fill the code parameter with code or use addition attributes to refer a code zipped and uploaded to S3.
E.g.
Code: { /* required */
S3Bucket: 'STRING_VALUE',
S3Key: 'STRING_VALUE',
S3ObjectVersion: 'STRING_VALUE',
ZipFile: new Buffer('...') || 'STRING_VALUE'
},
Also make sure to give required permission to the IAM User and setup JavaScript SDK credentials to run the code.