我有一个 RabbitMQ 消费者。我想让该消费者进行一些消息处理,由 模拟time.sleep(10)
,然后将消息发布到不同的队列。我知道消费者回调有一个理论上可以用来进行发布的通道,但这似乎是一个糟糕的实现,因为如果以basic_publish()
某种方式设法强制关闭通道,那么消费者就会死亡。处理这个问题的最佳方法是什么?
import time
import pika
connection = pika.BlockingConnection(
pika.ConnectionParameters(host='localhost'))
channel = connection.channel()
channel.exchange_declare(exchange='logs', exchange_type='fanout')
result = channel.queue_declare(queue='original_queue', exclusive=True)
channel.queue_bind(exchange='logs', queue='original_queue')
print(' [*] Waiting for logs. To exit press CTRL+C')
def callback(ch, method, properties, body):
time.sleep(10)
ch.basic_publish(exchange='logs', routing_key='different_queue', body='hello_world')
channel.basic_consume(
queue='original_queue', on_message_callback=callback, auto_ack=True)
channel.start_consuming()