[GitHub] dongeforever opened a new issue #419: 全局排序的Consumer

[GitBox <gi...@apache.org>]

dongeforever opened a new issue #419: 全局排序的Consumer
URL: https://github.com/apache/rocketmq/issues/419
 
 
   RocketMQ的Topic的分队列存储的，队列的特点如下：
   1.单个队列内，按照发送时间进行排序；
   2.队列之间无序。
   
   有些场景，比如股票撮合，需要全局对多个队列进行排序。
   
   可以考虑在客户端实现一个全局排序的OrderedConsumer
   `
   class OrderedConsumer{
           //按照发送时间（getBornTimestamp）进行排序
          MessageExt receive();
   }
   `
   可以利用PullConsumer进行实现：
   ` Set<MessageQueue> mqs = consumer.fetchSubscribeMessageQueues("TopicTest1");
           for (MessageQueue mq : mqs) {
               System.out.printf("Consume from the queue: %s%n", mq);
               SINGLE_MQ:
               while (true) {
                   try {
                       PullResult pullResult =
                           consumer.pullBlockIfNotFound(mq, null, getMessageQueueOffset(mq), 32);
                       System.out.printf("%s%n", pullResult);
                       putMessageQueueOffset(mq, pullResult.getNextBeginOffset());
                       switch (pullResult.getPullStatus()) {
                           case FOUND:
                               break;
                           case NO_MATCHED_MSG:
                               break;
                           case NO_NEW_MSG:
                               break SINGLE_MQ;
                           case OFFSET_ILLEGAL:
                               break;
                           default:
                               break;
                       }
                   } catch (Exception e) {
                       e.printStackTrace();
                   }
               }
           }
   `
   
   实现时，可以逐步来，先假定MessageQueue的数量是固定。
   
   

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