如何重命名 python 脚本以用于 lambda - lambda_handler vs main() function 名称？

Question

最初这是 ec2 上的 python 脚本，但现在我希望它成为 aws lambda - 从 terraform 生成。因为 aws lambda 需要 lambda_handler function 与“__ main __”。

我想知道在处理程序 arg 的 tf 代码中放入什么。

python 脚本（它被 terraform 压缩然后加载到 aws lambda）：

#!/usr/bin/env python3

import boto3
import json
import logging
import sys

logging.basicConfig(stream=sys.stdout, level=logging.INFO)

queue = boto3.resource(
    'sqs', region_name='us-east-1').get_queue_by_name(QueueName="erjan")
table = boto3.resource('dynamodb', region_name='us-east-1').Table('Votes')

def process_message(message):
    try:
        payload = message.message_attributes
        voter = payload['voter']['StringValue']
        vote = payload['vote']['StringValue']
        logging.info("Voter: %s, Vote: %s", voter, vote)
        update_count(vote)
        message.delete()
    except Exception as e:
        print('-----EXCEPTION-----')


def update_count(vote):
    logging.info('update count....')
    cur_count = 0
    if vote == 'b':
        response = table.get_item(Key={'voter': 'count'})
        item = response['Item']
        item['b'] += 1
        table.put_item(Item=item)

    elif vote == 'a':
        table.update_item(
            Key={'voter': 'count'},
            UpdateExpression="ADD a :incr",
            ExpressionAttributeValues={':incr': 1})


if __name__ == "__main__":

    logging.info('--------inside main-------')

    while True:
        try:
            messages = queue.receive_messages(MessageAttributeNames=['vote', 'voter'])
        except KeyboardInterrupt:
            logging.info("Stopping...")
            break
        except:
            logging.error(sys.exc_info()[0])
            continue
        for message in messages:
            process_message(message)

tf 代码：

resource "aws_iam_role" "vote_processor_lambda_iam_role" {
  name               = "vote_processor_lambda_iam_role"
  assume_role_policy = <<EOF
    {
        "Version": "2012-10-17",
        "Statement": [
            {
                "Action": "sts:AssumeRole",
                "Principal": {
                    "Service": "lambda.amazonaws.com"
                },
                "Effect": "Allow",
                "Sid": ""
            }
        ]
    }
    EOF
}


resource "aws_iam_policy" "vote_processor_dynamodb_policy" {

  name   = "vote_processor_dynamodb_policy"
  policy = <<EOF
    {
    "some json"
}
}


resource "aws_iam_role_policy_attachment" "attach_vote_processor_dynamodb_policy_to_iam_role" {
  role       = aws_iam_role.vote_processor_lambda_iam_role.name
  policy_arn = aws_iam_policy.vote_processor_dynamodb_policy.arn
}


resource "aws_iam_policy" "vote_processor_sqs_policy" {

 name   = "vote_processor_sqs_policy"
  policy = <<EOF
{
    "some json"
}
EOF
}

resource "aws_iam_role_policy_attachment" "vote_processor_sqs_access_policy" {
  role       = aws_iam_role.vote_processor_lambda_iam_role.name
  policy_arn = aws_iam_policy.vote_processor_sqs_policy.arn
}



data "archive_file" "vote_processor_zip_code" {
  type        = "zip"
  source_file = "${path.module}/vote_processor.py"
  output_path = "${path.module}/vote_processor.zip"
}



#what to put in handler arg? 

resource "aws_lambda_function" "vote_processor_lambda_backend" {
  filename      = "${path.module}/vote_processor.zip"
  function_name = "vote_processor"
  role          = aws_iam_role.vote_processor_lambda_iam_role.arn
  handler       = "result.lambda_handler"   #should this be __main__?
  runtime       = "python3.9"

}

我应该将 python 脚本main function 重命名为“lambda 处理程序”吗？ 或者在 tf 代码中反之亦然？

Answer 1

“__ main __”不是必需的。 以下语法定义了 lambda 个函数。

Lambda function 处理程序在 Python - AWS Lambda

在 Python 中创建 function 处理程序时，您可以使用以下通用语法：

 def handler_name(event, context): ... return some_value

为 tf 设置的处理程序如下。

Lambda function 处理程序在 Python - AWS Lambda

在您创建 Lambda function 时指定的 Lambda function 处理程序名称派生自：

Lambda 处理程序 function 所在文件的名称。

Python 处理程序 function 的名称。

function 处理程序可以是任何名称； 但是，Lambda 控制台中的默认名称是 lambda_function.lambda_handler。 这个 function 处理程序名称反映了 function 名称 (lambda_handler) 和存储处理程序代码的文件 (lambda_function.py)。