等待一个RabbitMQ消息超时 [英] Wait for a single RabbitMQ message with a timeout

问题描述

我想向RabbitMQ服务器发送一条消息，然后等待答复消息（在答复队列中）。当然，我不想永远等待，以防处理这些消息的应用程序关闭-需要超时。听起来这是一项非常基本的任务，但我找不到解决方法。我现在同时遇到 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.

我最好的解决方案到目前为止，已经使用 basic_get 和 sleep 进行了轮询，但这很丑陋：

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

我刚刚添加了对 amqplib 放在胡萝卜中。

I just added timeout support for amqplib in carrot.

这是 amqplib.client0_8.Connection ：

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

wait_multi 是 channel.wait 能够接收任意数量的
个频道。

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

我想这可能会在上游合并

I guess this could be merged upstream at some point.

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