没有“到”端点的 Apache Camel 路由

Question

我正在使用 Apache Camel 来协助捕获第三方软件包发出的消息数据。在这个特定的例子中，我只需要捕捉软件产生的东西，另一端没有接收器（真的没有“端”去）。

因此，我尝试设置一条只有“from”端点而没有“to”端点的路由。显然这是不正确的用法，因为我收到以下异常：

[2018-08-15 11:08:03.205] ERROR: string.Launcher:191 - Exception
org.apache.camel.FailedToCreateRouteException: Failed to create route route1 at: >>> From[mina:udp://localhost:9877?sync=false] <<< in route: Route(route1)[[From[mina:udp://localhost:9877?sync=false]] -... because of Route route1 has no output processors. You need to add outputs to the route such as to("log:foo").
    at org.apache.camel.model.RouteDefinition.addRoutes(RouteDefinition.java:1063)
    at org.apache.camel.model.RouteDefinition.addRoutes(RouteDefinition.java:196)
    at org.apache.camel.impl.DefaultCamelContext.startRoute(DefaultCamelContext.java:974)
    at org.apache.camel.impl.DefaultCamelContext.startRouteDefinitions(DefaultCamelContext.java:3301)
    at org.apache.camel.impl.DefaultCamelContext.doStartCamel(DefaultCamelContext.java:3024)
    at org.apache.camel.impl.DefaultCamelContext.access$000(DefaultCamelContext.java:175)
    at org.apache.camel.impl.DefaultCamelContext$2.call(DefaultCamelContext.java:2854)
    at org.apache.camel.impl.DefaultCamelContext$2.call(DefaultCamelContext.java:2850)
    at org.apache.camel.impl.DefaultCamelContext.doWithDefinedClassLoader(DefaultCamelContext.java:2873)
    at org.apache.camel.impl.DefaultCamelContext.doStart(DefaultCamelContext.java:2850)
    at org.apache.camel.support.ServiceSupport.start(ServiceSupport.java:61)
    at org.apache.camel.impl.DefaultCamelContext.start(DefaultCamelContext.java:2819)
    at {removed}.Launcher.startCamel(Launcher.java:189)
    at {removed}.Launcher.main(Launcher.java:125)
Caused by: java.lang.IllegalArgumentException: Route route1 has no output processors. You need to add outputs to the route such as to("log:foo").
    at org.apache.camel.model.RouteDefinition.addRoutes(RouteDefinition.java:1061)
    ... 13 more

如何设置允许我拦截（捕获）来自源的消息流量，而不是将其“发送到”任何东西的骆驼路由？不需要接收器。什么是合适的“to”端点，只是丢弃它接收到的所有内容？

的例外建议to("log:foo")。这有什么作用？

Answer 1

您可以看看该Stub组件是否可以提供帮助

http://camel.apache.org/stub.html

例子：

from("...")
    .to("stub:nowhere");