Question

我在python上运行代码，从另一个我不能允许线程的应用程序发送和接收RabbitMQ队列。这是一个非常新手的问题，但有没有可能只检查是否有消息，如果没有任何东西，那么只是退出听？我该如何更改此类任务的基本“Hello world”示例？目前我已经设法停止消费，如果我收到消息，但如果没有消息，我的方法接收（）只是继续等待。如果没有消息，如何强迫它不要等待？或者也许只等待一段时间？

import pika

global answer

def send(msg):
    connection = pika.BlockingConnection(pika.ConnectionParameters())
    channel = connection.channel()
    channel.queue_declare(queue='toJ')
    channel.basic_publish(exchange='', routing_key='toJ', body=msg)
    connection.close()

def receive():
    connection = pika.BlockingConnection(pika.ConnectionParameters(host='localhost'))
    channel = connection.channel()
    channel.queue_declare(queue='toM')
    channel.basic_consume(callback, queue='toM', no_ack=True)
    global answer
    return answer

def callback(ch, method, properties, body):
    ch.stop_consuming()
    global answer
    answer = body

Answer 1

好的，我找到了以下解决方案：

def receive():
    parameters = pika.ConnectionParameters(RabbitMQ_server)
    connection = pika.BlockingConnection(parameters)
    channel = connection.channel()
    channel.queue_declare(queue='toM')
    method_frame, header_frame, body = channel.basic_get(queue = 'toM')        
    if method_frame.NAME == 'Basic.GetEmpty':
        connection.close()
        return ''
    else:            
        channel.basic_ack(delivery_tag=method_frame.delivery_tag)
        connection.close() 
        return body

如果存在则RabbitMQ消耗一条消息并退出

1 个答案: