将RabbitMQ与Bunny(Ruby)一起使用,我想发布并希望避免在扇出交换中没有队列消耗时丢失消息。从我的图书馆角度来看,我愿意等待回复,并在发送消息时给客户一个明确的答案。在这方面睡觉是实现这一目标的唯一途径吗?
def publish(topic, message_type, message, key = nil)
ch = @conn.create_channel
exchange = ch.fanout(topic, :durable => true, :auto_delete => false)
sent = true
exchange.on_return do |return_info, properties, content|
sent = false
end
exchange.publish(message,
:mandatory => true,
:persistent => true,
:type => message_type,
:app_id => Emque::Producing.configuration.app_name,
:content_type => "application/json")
# Give message a chance to return if it can't be routed
sleep 0.5
ch.close
return sent
end
答案 0 :(得分:0)
basic.return是协议中的异步操作。你可以使用同步原语(例如Bunny有Concurrent :: Condition,因为Ruby缺少与java.util.concurrent类似的东西),但你真正想要的是发布者确认。请参阅“何时确认消息?”在http://www.rabbitmq.com/confirms.html。