在 python/pika 中消费多个队列

2022-01-11 00:00:00 python rabbitmq pika

问题描述

我正在尝试创建一个订阅多个队列的消费者，然后在消息到达时对其进行处理.

I am trying to create a consumer that would subscribe to multiple queues, and then process messages as they arrive.

问题是，当第一个队列中已经存在一些数据时，它会消耗第一个队列，而永远不会去消耗第二个队列.但是，当第一个队列为空时，它确实会转到下一个队列，然后同时消耗两个队列.

The problem is that when there is some data already present in the first queue, it consumes the first queue and never goes to consume the second queue. However, when the first queue is empty, it does go to the next queue, and then consumes both queues simultaneously.

我首先实现了线程，但想避开它，当 pika 库为我完成它时没有太多复杂性.以下是我的代码:

I had first implemented threading but want to steer clear of it, when pika library does it for me without much complexity. Below is my code:

import pika

mq_connection = pika.BlockingConnection(pika.ConnectionParameters('x.x.x.x'))
mq_channel = mq_connection.channel()
mq_channel.basic_qos(prefetch_count=1)


def callback(ch, method, properties, body):
    print body
    mq_channel.basic_ack(delivery_tag=method.delivery_tag)

mq_channel.basic_consume(callback, queue='queue1', consumer_tag="ctag1.0")
mq_channel.basic_consume(callback, queue='queue2', consumer_tag="ctag2.0")
mq_channel.start_consuming()

解决方案

一种可能的解决方案是使用非阻塞连接并消费消息.

One possible solution is to use non blocking connection and consume messages.

import pika


def callback(channel, method, properties, body):
    print(body)
    channel.basic_ack(delivery_tag=method.delivery_tag)


def on_open(connection):
    connection.channel(on_channel_open)


def on_channel_open(channel):
    channel.basic_consume(callback, queue='queue1')
    channel.basic_consume(callback, queue='queue2')


parameters = pika.URLParameters('amqp://guest:guest@localhost:5672/%2F')
connection = pika.SelectConnection(parameters=parameters,
                                   on_open_callback=on_open)

try:
    connection.ioloop.start()
except KeyboardInterrupt:
    connection.close()

这将连接到多个队列并相应地使用消息.

This will connect to multiple queues and will consume messages accordingly.

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