Question

如何关闭连接到扇出交换的Kombu ConsumerMixin队列，以便在我的消费者不活动时不从发布者累积数据？

我在Python 2.7中使用Kombu 3.0.24（带有RabbitMQ）。

以下是我的两个类的代码。我希望这些是相当通用的类，所以我可以将它们重用于直接队列和类似RPC的查询/回复。

问题是，如果我停止并重新启动使用者，旧数据将在消费者队列中等待我。我假设这是因为我需要在停止消费者时删除队列，但我无法弄清楚如何。感谢。

MessageConsumer.py

from kombu.mixins import ConsumerMixin
from kombu import Queue, Exchange, Connection
import logging

class MessageConsumer(ConsumerMixin):

    def __init__(self,
                 broker='amqp://',
                 exchange='mExchange',
                 queue = 'mQueue',
                 type='direct',
                 no_ack=False):
        self.connection = Connection(broker)
        self.mExchange = Exchange(exchange, type=type)
        self.mQueue = Queue(queue, self.mExchange)
        self.mQueue.no_ack = no_ack

    def get_consumers(self, Consumer, channel):
        return [Consumer(queues=self.mQueue,
                         accept=['json'],
                         callbacks=[self.process_task])]

    def process_task(self, body, message):
        logging.debug('RECEIVED: {}'.format(body))

    def stop(self):
        self.should_stop = True
        self.connection.release()


if __name__ == '__main__':

    mMessageConsumer = MessageConsumer(exchange='sensor_data',
                                       queue='rx1_queue',
                                       type='fanout',
                                       no_ack=True)
    try:
        mMessageConsumer.run()
    except KeyboardInterrupt:
        mMessageConsumer.stop()

MessagePublisher.py

from kombu import Queue, Exchange, Connection
from kombu.pools import producers
import logging

class MessagePublisher(object):

    def __init__(self,
                 broker='amqp://',
                 exchange='mExchange',
                 type='direct',
                 no_ack=False):
        self.connection = Connection(broker)
        self.mExchange = Exchange(exchange, type=type)

    def publish(self, message, serializer='json', compression=None):
        with producers[self.connection].acquire(block=True) as producer:
            producer.publish(message,
                             serializer=serializer,
                             compression=compression,
                             exchange=self.mExchange,
                             declare=[self.mExchange]
                             )

    def close(self):
        self.connection.release()

if __name__ == '__main__':
    mMessagePublisher = MessagePublisher(type='fanout',exchange='sensor_data')
    x=0
    while True:
        x += 1
        mMessagePublisher.publish(x)
    mMessagePublisher.close()

如果我有更有效的方法对此进行编码，请提出建议。我通过谷歌搜索找到的大部分例子都使用了较早版本的Kombu，所以我很难找到3.0.24的最佳实现。

Answer 1

找到解决方案。我需要使用＆＃34; exclusive = True＆＃34;创建我的队列。然后队列仅在我的程序使用时才存在。具体做法是：

self.mQueue = Queue(queue, self.mExchange, exclusive=True)

Kombu ConsumerMixin关闭队列以防止扇出数据接收

1 个答案: