RabbitMQ consume one message if exists and quit

Question

I am running code on python to send and receive from RabbitMQ queue from another application where I can't allow threading. This is very newbie question but, is there a possibility to just check if there is message and if there are no any then just quit listening ? How should I change basic "Hello world" example for such task? Currently I've managed to stop consuming if I get a message, but if there are no messages my method receive() just continue waiting. How to force it not to wait if there are no messages? Or maybe wait only for given amount of time?

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

Ok, I found following solution:

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