SQS receive_message 的正确解析是什么？

Question

几天来我一直在解决这个问题，但我似乎不知道该去哪里解决这个错误。我已经根据 Boto3、SQS 文档和 Lambda 测试中记录的内容对 receive_message 响应语法进行了测试，但无济于事。

{
      "errorMessage": "'Messages'",
      "errorType": "KeyError",
      "stackTrace": [
        "  File \"/var/task/lambda_function.py\", line 24, in lambda_handler\n    message = response['Messages'][0]\n"
      ]
 }

我直接从 Boto3 SDK documentation:

中截取了这个 Python 片段

import boto3
import os

# Create SQS client
sqs = boto3.client('sqs')

QUEUE_URL = os.environ['QUEUE_URL']

def lambda_handler(event, context):
    # Receive message from SQS queue
    response = sqs.receive_message(
        QueueUrl=QUEUE_URL,
        AttributeNames=[
            'SentTimestamp'
        ],
        MaxNumberOfMessages=10,
        MessageAttributeNames=[
            'All'
        ],
        VisibilityTimeout=0,
        WaitTimeSeconds=0
    )
    
    message = response['Messages'][0]
    receipt_handle = message['ReceiptHandle']
    
    # Delete received message from queue
    sqs.delete_message(
        QueueUrl=queue_url,
        ReceiptHandle=receipt_handle
    )
    print('Received and deleted message: %s' % message)

Answer 1

如果您配置了 Amazon SQS 来触发 AWS Lambda 函数，那么您不需要调用任何 SQS 命令。相反，消息在传递给 Lambda 函数的 event 中提供：

for record in event['Records']:
    payload = record['body']
    print(payload)

如果 Lambda 函数成功退出，消息将自动从队列中删除。

参见：Using Lambda with Amazon SQS - AWS Lambda

SQS receive_message 的正确解析是什么？

What is the proper parsing for SQS receive_message?

python

amazon-web-services

boto3

amazon-sqs