apache-camel - Camel-拆分列表并处理每个java对象-XML配置

Question

我无法拆分列表中的 Java 对象。如何将 Body 标记化/转换为单个 Java 对象？

    <route id="cleanupMigratedFiles" autoStartup="true">

        <from uri="timer://kickoff?period=5s" />
        <bean ref="migrationProcessor" method="getCacheDeleteObjects" /> <!--  this gives me a List-of-CacheMigr -->
        <log message="\n\t########\n\tCleanupMigrated file: ${body}" />
        <pipeline>
            <split>
                <tokenize /> <!-- How to tokenize a List-of-CacheMigr  -->
                <convertBodyTo type="era.oddw.entity.CacheMigr" /> <!-- Do I need this? -->
                <log message="\n\t########\n\tCleanupMigrated file: ${body}" />
            </split>
        </pipeline>
    </route>

Answer 1

经过更多阅读后找到了答案..以下标记正确标记列表:${body}

感谢 Camel 人。

        <split streaming="true">
            <simple>${body}</simple>
            <convertBodyTo type="era.oddw.entity.CacheMigr" /> 
            <log message="\n\t########\n\tCleanupMigrated file each: ${body}" />
        </split>