Spring Integration:基于内容的路由器,默认输出通道？

Question

我想使用Spring Integration来实现基于内容的路由器,如果表达式值与任何映射都不匹配,则使用默认输出通道.这是我的bean定义:

<int:router input-channel="channel_in" default-output-channel="channel_default" expression="payload.name">
    <int:mapping value="foo" channel="channel_one" />
    <int:mapping value="bar" channel="channel_two" />

但是,似乎从不使用默认输出通道.如果表达式评估为例如'baz',则路由器似乎正在寻找名为'baz'的频道,而不是路由到'channel_default'频道:

org.springframework.integration.MessagingException: failed to resolve channel name 'baz'
  Caused by: org.springframework.integration.support.channel.ChannelResolutionException: 
    failed to look up MessageChannel bean with name 'baz'
  Caused by: org.springframework.beans.factory.NoSuchBeanDefinitionException: 
    No bean named 'baz' is defined

我想要的是使用XML命名空间,或者我是否需要编写自己的实现？

Answer 1

事实证明,我必须做的就是将路由器的ignore-channel-name-resolution-failures属性设置为false:

<int:router input-channel="channel_in" default-output-channel="channel_default" 
  expression="payload.name" ignore-channel-name-resolution-failures="true">
    <int:mapping value="foo" channel="channel_one" />
    <int:mapping value="bar" channel="channel_two" />
</int:router>

我以为我之前尝试过,但我似乎没有.