是否可以在 Python 中将 RabbitMQ 直接回复功能与 Pika 生成器消费者一起使用？

Question

我想在 Python 中使用 direct reply-to feature of RabbitMQ with the Pika 客户端库。它适用于 basic 消费者。但它引发了以下异常 generator consumer:

pika.exceptions.ChannelClosedByBroker: (406, 'PRECONDITION_FAILED - fast reply consumer does not exist')

有没有办法使用生成器消费者的直接回复功能？

使用基本消费者的示例客户端代码（有效）：

import pika


def handle(channel, method, properties, body):
    message = body.decode()
    print("received:", message)


connection = pika.BlockingConnection()
channel = connection.channel()

with connection, channel:
    message = "hello"
    channel.basic_consume(queue="amq.rabbitmq.reply-to",
                          on_message_callback=handle, auto_ack=True)
    channel.basic_publish(
        exchange="", routing_key="test", body=message.encode(),
        properties=pika.BasicProperties(reply_to="amq.rabbitmq.reply-to"))
    print("sent:", message)
    channel.start_consuming()

使用生成器消费者的示例客户端代码（引发异常）：

import pika


def handle(channel, method, properties, body):
    message = body.decode()
    print("received:", message)


connection = pika.BlockingConnection()
channel = connection.channel()

with connection, channel:
    message = "hello"
    channel.basic_publish(
        exchange="", routing_key="test", body=message.encode(),
        properties=pika.BasicProperties(reply_to="amq.rabbitmq.reply-to"))
    print("sent:", message)

    for (method, properties, body) in channel.consume(
            queue="amq.rabbitmq.reply-to", auto_ack=True):
        handle(channel, method, properties, body)

环境。 — Windows 10，RabbitMQ 3.7.13，CPython 3.7.3，鼠兔 1.0.1。

注意。 — 在示例客户端代码中使用basic_publish方法后调用basic_consume方法基本消费者引发与使用生成器消费者时相同的异常：

import pika


def handle(channel, method, properties, body):
    message = body.decode()
    print("received:", message)


connection = pika.BlockingConnection()
channel = connection.channel()

with connection, channel:
    message = "hello"
    channel.basic_publish(
        exchange="", routing_key="test", body=message.encode(),
        properties=pika.BasicProperties(reply_to="amq.rabbitmq.reply-to"))
    print("sent:", message)
    channel.basic_consume(queue="amq.rabbitmq.reply-to",
                          on_message_callback=handle, auto_ack=True)
    channel.start_consuming()

Answer 1

正如 Luke Bakken here 所建议的那样，这样做就可以了：

import pika


def handle(channel, method, properties, body):
    message = body.decode()
    print("received:", message)


connection = pika.BlockingConnection()
channel = connection.channel()

with connection, channel:
    message = "hello"
    next(channel.consume(queue="amq.rabbitmq.reply-to", auto_ack=True,
                         inactivity_timeout=0.1))
    channel.basic_publish(
        exchange="", routing_key="test", body=message.encode(),
        properties=pika.BasicProperties(reply_to="amq.rabbitmq.reply-to"))
    print("sent:", message)

    for (method, properties, body) in channel.consume(
            queue="amq.rabbitmq.reply-to", auto_ack=True):
        handle(channel, method, properties, body)

Answer 2

我会推荐 pika 文档中给出的示例。

它对我也有用。

this is the location

是否可以在 Python 中将 RabbitMQ 直接回复功能与 Pika 生成器消费者一起使用？

Is it possible to use RabbitMQ direct reply-to feature with a Pika generator consumer in Python?

python

generator

rabbitmq

consumer

pika