[rabbitmq-discuss] Publish on a queue, then Monitor on a reply queue, with only 1 connection to the RabbitMQ server? Non-blocking?

[Christopher Lefevre]

To note, if this is an easier description, I’m trying to implement the RPC example on the RabbitMQ tutorial with a synchronous connection. Which I’ve copied below so you guys don’t need to go find it. However, the main impetus to me moving forward is lines 13/14:

        result = self.channel.queue_declare(exclusive=True)
        self.callback_queue = result.method.queue



Line 13 declares the result variable as the product of the channel getting an exclusive queue created, which I get. However the part that I get caught up in is where the callback is set to result.method.queue in line 14. What exactly is the method.queue of the result object? I am hoping to simply modify my Monitor/Publish classes to be able to have a synchronous publish/monitor without creating 2 connections. My current attempt at simply using one connection with 2 channels does not seem to be operating how I expected.

#!/usr/bin/env python
import pika
import uuid

class FibonacciRpcClient(object):
    def __init__(self):
        self.connection = pika.BlockingConnection(pika.ConnectionParameters(
                host='localhost'))

        self.channel = self.connection.channel()

        result = self.channel.queue_declare(exclusive=True)
        self.callback_queue = result.method.queue

        self.channel.basic_consume(self.on_response, no_ack=True,
                                   queue=self.callback_queue)

    def on_response(self, ch, method, props, body):
        if self.corr_id == props.correlation_id:
            self.response = body

    def call(self, n):
        self.response = None
        self.corr_id = str(uuid.uuid4())
        self.channel.basic_publish(exchange='',
                                   routing_key='rpc_queue',
                                   properties=pika.BasicProperties(
                                         reply_to = self.callback_queue,
                                         correlation_id = self.corr_id,
                                         ),
                                   body=str(n))
        while self.response is None:
            self.connection.process_data_events()
        return int(self.response)

fibonacci_rpc = FibonacciRpcClient()

print " [x] Requesting fib(30)"
response = fibonacci_rpc.call(30)
print " [.] Got %r" % (response,)



-Christopher Lefevre

From: Michael Klishin [mailto:michael.s.klishin at gmail.com]
Sent: Tuesday, September 06, 2011 9:38 AM
To: Christopher Lefevre
Cc: rabbitmq-discuss at lists.rabbitmq.com
Subject: Re: [rabbitmq-discuss] Publish on a queue, then Monitor on a reply queue, with only 1 connection to the RabbitMQ server? Non-blocking?

2011/9/6 Christopher Lefevre <CLefevre at transparent.com<mailto:CLefevre at transparent.com>>
Though I also wanted it to be non-blocking so the website isn’t stuck waiting for the reply, but rather can be updated as needed. Currently, I can get it to work if there is a pause, like when stepping through the program in a debugger, but when it runs without myself stepping through, it simply hangs when the queue is created and the message is never picked up.

Please post a code example that is at least similar to your actual code.

--
MK

http://github.com/michaelklishin
http://twitter.com/michaelklishin
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