如何丰富mule中的xml节点值[重复]

Question

input xml like below
<healthcare>
    <plans>
        <plan1>
            <planid>100</planid>
            <planname>medical</planname>
            <desc>medical</desc>
            <offerprice>3000</offerprice>
            <area>texas</area>
        </plan1>
        <plan2>
            <planid>101</planid>
            <planname>dental</planname>
            <desc>dental</desc>
            <offerprice>4000</offerprice>
            <area>texas</area>
        </plan2>
    </plans>
</healthcare>



 <set-variable variableName="newoffer" value="2000" doc:name="Variable" />
<foreach doc:name="For Each" collection="#[xpath3('//healthcare/plans', message.payload, 'NODESET' )]">
<enricher source="#[flowVars.newoffer]" target="#[xpath3('offerprice', payload 'STRING')]" >
<logger message="...#[message.payloadAs(java.lang.String)]" level="INFO" doc:name="Logger"/>
</enricher>
<logger message="#[message.payloadAs(java.lang.String)]" level="INFO" doc:name="Logger"/>
</foreach>

我想将 xml 元素的 offerprice 值丰富为 2000 。但上面没有这样做。任何人有任何建议。主要是我的预期输出如下所示。我是 mule 新手。任何人都可以了解我的 mule 需求。提前致谢

<healthcare>
    <plans>
        <plan1>
            <planid>100</planid>
            <planname>medical</planname>
            <desc>medical</desc>
            **<offerprice>2000</offerprice>**
            <area>texas</area>
        </plan1>
        <plan2>
            <planid>101</planid>
            <planname>dental</planname>
            <desc>dental</desc>
            **<offerprice>2000</offerprice>**
            <area>texas</area>
        </plan2>
    </plans>
</healthcare>

Answer 1

要实现您的目标，请删除 for-each / enricher 元素，而改用 Mule 附带的 XSLT Transformer。这是将 XML 转换为 XML 的最佳选择。