使用 kombu 为 rabbitmq 实现 RPC 客户端及其直接回复功能

Question

我正在实现一个带有 kombu 的 rpc 框架，以便与兔子一起使用我想使用 rabbitmq 的直接回复功能，但我找不到使用 kombu 实现它的方法。

客户端在生成消息之前需要在"amq.rabbitmq.reply-to"队列上使用，并且生产者和消费者应使用相同的通道。我还需要使用生产者池(或某种连接池(，因为客户端是在线程环境中创建的。

到目前为止，我有这段代码，rabbitmq 不会抱怨没有生产者出现 PRECONDITION 错误(如果我删除消费者部分，它确实会抱怨(，但生产者不会生产任何东西！

class KombuRpcClient(RpcClientBase):
    def __init__(self, params):
        self.future = Queue.Queue()
        self.logger = logger
        if isinstance(params, RpcConnectionProperties):
            self.rpc_connection_properties = params
        else:
            self.rpc_connection_properties = RpcConnectionProperties(
                host=params.get('host'),
                port=5672,
                username=params.get('username'),
                password=params.get('password'),
                vhost=params.get('vhost') if params.has_key('vhost') else '/'
            )
        self.amqp_url = self.rpc_connection_properties.get_kombu_transport_url('pyamqp')
        self.reply_queue = KombuQueue('direct_reply', exchange=default_exchange, routing_key='amq.rabbitmq.reply-to')
    def call(self, exchange, key, msg, no_response=False, timeout=5):
        connection = Connection(self.amqp_url)
        if exchange is not None:
            key = exchange + ':' + key
        with producers_pool[connection].acquire(block=True) as producer:
            with producer.channel.Consumer(queues=[self.reply_queue], no_ack=True, callbacks=[self._on_message],
                                           accept=['ujson']) as consumer:
                producer.publish(
                    msg,
                    exchange=default_exchange,
                    routing_key=key,
                    immediate=True,
                    serializer='ujson', reply_to=self.reply_queue.routing_key)
                consumer.consume()
                pass
        res = self.future.get(block=True, timeout=timeout)
        print res
    def cast(self, exchange, key, msg):
        pass
    def _on_message(self, body, message):
        print body
        self.future.put(body)

Answer 1

在wireshark的帮助下，我意识到有时rabbitmq会响应该PRECONDITION错误，但kombu不会引发异常，我不知道为什么！无论如何，这段代码现在可以工作了：

class KombuRpcClient(RpcClientBase):
    def __init__(self, params):
        self.future = Queue.Queue()
        self.logger = logger
        if isinstance(params, RpcConnectionProperties):
            self.rpc_connection_properties = params
        else:
            self.rpc_connection_properties = RpcConnectionProperties(
                host=params.get('host'),
                port=5672,
                username=params.get('username'),
                password=params.get('password'),
                vhost=params.get('vhost') if params.has_key('vhost') else '/'
            )
        self.amqp_url = self.rpc_connection_properties.get_kombu_transport_url('pyamqp')
        self.reply_queue = KombuQueue('amq.rabbitmq.reply-to', exchange=default_exchange, routing_key='amq.rabbitmq.reply-to')
    def call(self, exchange, key, msg, no_response=False, timeout=5):
        connection = Connection(self.amqp_url)
        if exchange is not None:
            key = exchange + ':' + key
        with producers_pool[connection].acquire(block=True) as producer:
            consumer = producer.channel.Consumer(queues=[self.reply_queue], no_ack=True, auto_declare=True,
                                callbacks=[self._on_message], accept=['ujson'])
            consumer.consume(no_ack=True)
            producer.publish(msg,
                             serializer='ujson',
                             exchange=default_exchange,
                             routing_key=key,
                             reply_to='amq.rabbitmq.reply-to')
            consumer.connection.drain_events()
            res = self.future.get(block=True, timeout=timeout)
            response = Response()
            response.body = res
            return res
    def cast(self, exchange, key, msg):
        connection = Connection(self.amqp_url)
        if exchange is not None:
            key = exchange + ':' + key
        with producers_pool[connection].acquire(block=True) as producer:
            producer.publish(msg,
                             serializer='ujson',
                             exchange=default_exchange,
                             routing_key=key)
    def _on_message(self, body, message):
        print body
        self.future.put(body)

顺便说一句：很抱歉变量名称，这应该替换旧的 stomp rpc 客户端，所以我不得不保留名称以兼容