Question

我根据有效载荷选择路线。有效负载可以是列表，映射或字符串。必须根据有效载荷类型识别和路由有效载荷。

Answer 1

我认为你想要那样的东西

<choice doc:name="Choice">
        <when expression="#[payload is List]">
            <logger level="INFO" message="i am list" doc:name="Logger"/>
        </when>
        <when expression="#[payload is Map]">
            <logger level="INFO" message="i am map" doc:name="Logger"/>
        </when>
        <when expression="#[payload is String]">
            <logger level="INFO" message="i am string" doc:name="Logger"/>
        </when>
        <otherwise> 
            <logger message="class doesnt match with [list, map, string]" level="INFO" doc:name="Logger"/>  
        </otherwise>  
    </choice>

Answer 2

我同意Eddú的回复，除了我会添加otherwise路径。如果你不这样做并得到一些不同于列表地图或字符串的东西，你会得到相当神秘的错误：

org.mule.api.routing.RoutePathNotFoundException: Can't process message because no route has been found matching any filter and no default route is defined. Failed to route event via endpoint: ChoiceRouter [flow-construct=testNull, started=true]. Message payload is of type: NullPayload

骡子选择路线取决于有效载荷类型，即对象类型？

2 个答案: