pika - Pika/RabbitMQ 如何从单个连接创建新通道？

Question

我有一个使用 python pica 的方法。这里消费者需要为每个队列创建一个通道。编写此代码时出现错误。

在此处输入图像描述

def Process(channel, exc, que):
    channel.exchange_declare(exchange=exc, exchange_type='direct', durable=True)
    result = channel.queue_declare(durable=True, queue=que, auto_delete=False,exclusive=False )

    def callback_rabbit(ch,method,properties,body):
        print("Message received = ", body)

    channel.queue_bind(exchange=exc, queue=que, routing_key=que)
    channel.basic_consume(on_message_callback=callback_rabbit,queue=que, auto_ack=True)
    channel.start_consuming()

def Start():        
    credentials = pika.PlainCredentials('guest','guest')
    connection = pika.BlockingConnection(pika.ConnectionParameters('127.0.0.1','5672','/',credentials))

    items = {"exc":[{"exchangeName":"bam", "queueName":"bam_queue1"},{"exchangeName":"bam", "queueName":"bam_queue2"},{"exchangeName":"bam", "queueName":"bam_queue3"}]}
    for item in items["exc"]:
        channel = connection.channel()
        t_msg = Thread(target=Process, args=(channel,item["exchangeName"],item["queueName"]))
        t_msg.start()

标签： pika

Answer 1

Pika 不是线程安全的。您应该在您的Process方法中创建一个新的连接和通道。

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