我的问题与之前提出的问题类似，但是它没有找到答案，我有一个消费者，我想处理一个名为Web服务的操作，但是，如果这个Web服务由于某种原因没有响应，我希望消费者不要处理RabbitMQ的消息，但我会让它稍后处理它，我的消费者是下面的一个：

require File.expand_path('../config/environment.rb', __FILE__)
conn=Rabbit.connect
conn.start
ch = conn.create_channel
x = ch.exchange("d_notification_ex", :type=> "x-delayed-message", :arguments=> { "x-delayed-type" => "direct"})
q  = ch.queue("d_notification_q", :durable =>true)
q.bind(x)
p 'Wait ....'
q.subscribe(:manual_ack => true, :block => true) do |delivery_info, properties, body|
 
  datos=JSON.parse(body)
  if datos['status']=='request'
    #I call a web service and process the json
    result=Notification.send_payment_notification(datos.to_json)
  else
    #I call a web service and process the body
    result=Notification.send_payment_notification(body)
  end
   #if the call to the web service, the web server is off the result will be equal to nil
   #therefore, he did not notify RabbitMQ, but he puts the message in UNACKED status
   # and does not process it later, when I want him to keep it in the queue and evaluate it afterwards.
  unless result.nil?
  ch.ack(delivery_info.delivery_tag)
  end

end

RabbitMQ的图片，

在声明中有某种方式：c hack（delivery_info.delivery_tag），这不是删除队列中的元素，可以稍后再处理，有什么想法吗？谢谢