等待一条带有超时的 RabbitMQ 消息

问题描述

我想向 RabbitMQ 服务器发送一条消息，然后等待回复消息(在回复"队列上).当然，我不想永远等待，以防处理这些消息的应用程序出现故障 - 需要超时.这听起来像是一项非常基本的任务，但我找不到这样做的方法.我现在遇到了 py-amqplib 和 RabbitMQ .NET 客户端.

I'd like to send a message to a RabbitMQ server and then wait for a reply message (on a "reply-to" queue). Of course, I don't want to wait forever in case the application processing these messages is down - there needs to be a timeout. It sounds like a very basic task, yet I can't find a way to do this. I've now run into this problem with both py-amqplib and the RabbitMQ .NET client.

到目前为止，我得到的最佳解决方案是使用 basic_get 和中间的 sleep 进行轮询，但这很丑:

The best solution I've got so far is to poll using basic_get with sleep in-between, but this is pretty ugly:

def _wait_for_message_with_timeout(channel, queue_name, timeout):
    slept = 0
    sleep_interval = 0.1

    while slept < timeout:
        reply = channel.basic_get(queue_name)
        if reply is not None:
            return reply

        time.sleep(sleep_interval)
        slept += sleep_interval

    raise Exception('Timeout (%g seconds) expired while waiting for an MQ response.' % timeout)

肯定有更好的方法吗?

推荐答案

我刚刚在 carrot 中添加了对 amqplib 的超时支持.

I just added timeout support for amqplib in carrot.

这是 amqplib.client0_8.Connection 的子类:

http://github.com/询问/carrot/blob/master/carrot/backends/pyamqplib.py#L19-97

wait_multi 是 channel.wait 的一个版本，能够接收任意数字频道.

wait_multi is a version of channel.wait able to receive on an arbitrary number of channels.

我想这可能会在某个时候在上游合并.

I guess this could be merged upstream at some point.

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