我想将我的同步API之一转换为异步。我相信队列是做到这一点的一种方法。就像出版商一样，（同步）将消息推入队列，该消息将由消费者API从队列中消费。

Question

如果您想从SQS消耗，则具有以下方法：

使用SDK填充以消费消息

使用Amazon SQS Java消息库

Answer 1

订阅SNS主题

• 使用lambda
.
• 如果您打算检索回复，您也可以利用
虚拟队列
• 。
在应用程序

• sqs:
  region: ap-south-1
  accessKeyId: arunsinghgujjar
  secretAccessKey: jainpurwalearunsingh/saharanpursepauchepuna
  cloud:
  aws:
  end-point:
  uri: https://arun-learningsubway-1.amazonaws.com/9876974864/learningsubway_SQS.fifo
  queue:
  max-poll-time: 20
  max-messages: 10
  fetch-wait-on-error: 60
  enabled: true
  content: sqs

写入SQS客户端

public String sendMessage(MessageDistributionEvent messageDistributionEvent) {
  SendMessageResponse sendMessage = null;
  try {
    Map<String, MessageAttributeValue> attributes = new HashMap<>();

    String recepList = "";
    for (Integer myInt : messageDistributionEvent.getRecipients()) {
      recepList = recepList + "_" + myInt;
    }
    SendMessageRequest sendMsgRequest = SendMessageRequest.builder()
      .queueUrl(url)
      .messageBody(messageDistributionEvent.getChannelId() + "_" + messageDistributionEvent.getMessageId() + "" + recepList)
      .messageGroupId("1")
      .messageAttributes(attributes)
      .build();
    sendMessage = sqsClient.sendMessage(sendMsgRequest);
  } catch (Exception ex) {
    log.info("failed to send message :" + ex);
  }
  return sendMessage.sequenceNumber();
}

从队列阅读消息

ReceiveMessageRequest receiveMessageRequest = ReceiveMessageRequest.builder()
.queueUrl(url)
.waitTimeSeconds(maxPollTime)
.maxNumberOfMessages(maxMessages)
.messageAttributeNames("MessageLabel")
.build();
List<Message> sqsMessages = sqsClient.receiveMessage(receiveMessageRequest).messages();

Answer 2

参考

https：//learningsubway.com/read-write-data-into-into-aws-sqs-using-java/