Question

我有一条路由收集标头，然后将请求发送到另一个队列，我从那里等待一个答案（我只需要响应主体）。问题是我想在一条路由中的一条消息中同时收到标头和答案，但是现在我在一条路由中得到了两条消息（我需要汇总响应正文和标头）。怎么做？

from("jms:queue:aaa")
    .log("incoming message")
    .process(exchange -> {
        exchange.getIn().setHeader("c", c);
        exchange.getIn().setHeader("d", d);
        exchange.getIn().setHeader("a", a);
        exchange.getIn().setHeader("b", b);
        exchange.getIn().setBody("2+3");
    })
    .removeHeaders("*", "a", "b", "c", "d")
    .setHeader("JMSReplyTo", simple("bbb"))
        //send request
    .to(ExchangePattern.InOnly, "jms:ccc?preserveMessageQos=true&includeSentJMSMessageID=true")
        //trying to send headers from this route to bbb
    .to(ExchangePattern.InOut, "jms:bbb")
.end();
from("jms:bbb")
    .log("${headers}\n${}body")
.end();

from（“ jms：bbb”）这是我要合并来自。
的结果的路线 .to(ExchangePattern.InOnly, "jms:ccc?preserveMessageQos=true& includeSentJMSMessageID=true")。
和。
.to (ExchangePattern.InOut, "jms: bbb")

UPD：

public class AggregationStrategyImpl implements AggregationStrategy {

    private static final Logger log = LoggerFactory.getLogger(AggregationStrategyImpl.class);

    @Override
    public Exchange aggregate(Exchange oldExchange, Exchange newExchange) {
        if(newExchange == null)
            newExchange = oldExchange;
        String a = oldExchange.getIn().getHeader("a", String.class);
        String b = oldExchange.getIn().getHeader("b", String.class);
        String c = oldExchange.getIn().getHeader("c", String.class);
        String d = oldExchange.getIn().getHeader("d", String.class);
        newExchange.getIn().setHeader("a", a);
        newExchange.getIn().setHeader("b", b);
        newExchange.getIn().setHeader("c", c);
        newExchange.getIn().setHeader("d", d);

        return newExchange;
    }
}

 .setHeader("JMSReplyTo", simple("bbb"))
                    .multicast(aggregationStrategy)
                    .to(ExchangePattern.InOnly, "jms:ccc?preserveMessageQos=true&includeSentJMSMessageID=true")
                    .to(ExchangePattern.OutOnly, "jms:bbb")

Answer 1

                .log("incoming message")
                .process(exchange -> {
                    exchange.getIn().setHeader("c", c);
                    exchange.getIn().setHeader("d", d);
                    exchange.getIn().setHeader("a", a);
                    exchange.getIn().setHeader("b", b);
                    exchange.getIn().setBody("2+3");
                })
                .removeHeaders("*", "a", "b", "c", "d")
                .setHeader("JMSReplyTo", simple("bbb"))
                //send request
                .multicast(aggregateStratergeyBean)
                .to(ExchangePattern.InOnly, "jms:ccc?preserveMessageQos=true&includeSentJMSMessageID=true")
                //trying to send headers from this route to bbb
                .to(ExchangePattern.InOut, "jms:bbb")
                .end();
        from("jms:bbb")
                .log("${headers}\n${}body")
                .end();```

aggregateStratergeyBean should implements AggregationStrategy. 

Also you can use many options like below

.parallelProcessing(true)   // Parallel process both routes

.parallelAggregate()  // Aggregate parallel , you need to handle thread safety

.streaming(). // Process response as and when it get the response not in the order it invoked.

Answer 2

请参见here。您的第一个if语句似乎无效：

if(newExchange == null) 
  newExchange = oldExchange;

这样，新传入的交换对象将始终保持为空。相反，应该是另一种方式，因为当oldExchange进入时，您想保留newExchange：

if (oldExchange == null) {
    //...
    return newExchange;
} else {
    //...
    return oldExchange;
}

问题似乎是您在等待newExchange为空，但是如果调用了聚合，则总是是newExchange。即将出现的第一个交换对象也是newExchange。相反，聚合的第一个调用没有oldExchange！

如何在一条路由中汇总两个消息？

2 个答案: