首页 \ 问答 \ Rabbitmq订阅多种服务,但以循环方式使用它(Rabbitmq Subscribe to multiple services but consume it in a round robin fashion)

Rabbitmq订阅多种服务,但以循环方式使用它(Rabbitmq Subscribe to multiple services but consume it in a round robin fashion)

我已经找到了解决问题的方法,但想知道是否有更清洁的方法。

我的架构由几个服务组成,通过Rabbitmq代理发送消息。

一些工人消费这些消息并做后台工作。

问题是我希望能够创建不同类型的工作人员,所有人都使用相同的服务,并且能够让几个相同类型的工人通过循环来完成工作。

为此,消息由服务以pub / sub方式发布,并由在专用于一组worker的工作队列中重新分发消息的进程使用。

这样做有更优雅的方式吗?

很抱歉,如果解释不清楚,我会编辑它。

谢谢!

(我可以在服务中为每个工作者创建一个队列,但是使用我的解决方案,我可以根据需要订阅,而无需触及任何代码)


I've got a working solution to my problem but wonder if there is a cleaner way of doing this.

My architecture is composed of several services, emitting messages through a Rabbitmq broker.

Some workers consume those messages and do background jobs.

The thing is that i wanted to be able to create different types of workers, all consuming the same services and be able to have several workers of the same type get the job through round robin.

To do this the messages are published by the service in a pub/sub fashion and consumed by a process that redistribute the messages in a work queue dedicated to a set of workers.

Is there a more elegant manner to do this?

Sorry if the explanation is not clear i'll edit it.

Thanks!

(I could have created one queue per worker in the services but with my solution I can subscribe as much as I want without touching any code)


原文:https://stackoverflow.com/questions/36551249
更新时间:2023-06-13 21:06

最满意答案

反序列化是读取XML文档并构造一个强类型为文档的XML Schema(XSD)的对象的过程。

你会做这样的事情

XmlSerializer serializer = new XmlSerializer(typeof(Items));

// Declare an object variable of the type to be deserialized.
Items i;

using (Stream reader = new FileStream(filename, FileMode.Open))
{
    // Call the Deserialize method to restore the object's state.
    i = (Items)serializer.Deserialize(reader);          
}

Deserialization is the process of reading an XML document and constructing an object that is strongly typed to the XML Schema (XSD) of the document.

You would do something like this

XmlSerializer serializer = new XmlSerializer(typeof(Items));

// Declare an object variable of the type to be deserialized.
Items i;

using (Stream reader = new FileStream(filename, FileMode.Open))
{
    // Call the Deserialize method to restore the object's state.
    i = (Items)serializer.Deserialize(reader);          
}

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