基于XSLT 3.0按applicationId及元素名分组去重XML的技术问询

问题背景 #

输入XML如下：

<?xml version='1.0' encoding='UTF-8'?>
<Applications>
    <Application>
        <applicationId>280</applicationId>
        <cust_IDCheckIDCollation>
            <Option>
                <id>197249</id>
            </Option>
        </cust_IDCheckIDCollation>
        <cust_overallduedilligencestatus>
            <Option>
                <id>197276</id>
            </Option>
        </cust_overallduedilligencestatus>
    </Application>
    <Application>
        <applicationId>292</applicationId>
        <cust_IDCheckIDCollation>
            <Option>
                <id>197249</id>
            </Option>
        </cust_IDCheckIDCollation>
        <cust_overallduedilligencestatus>
            <Option>
                <id>197276</id>
            </Option>
        </cust_overallduedilligencestatus>
    </Application>
    <Application>
        <applicationId>280</applicationId>
        <cust_OnlineReferenceCheck>
            <Option>
                <id>197249</id>
            </Option>
        </cust_OnlineReferenceCheck>
        <cust_overallduedilligencestatus>
            <Option>
                <id>197276</id>
            </Option>
        </cust_overallduedilligencestatus>
    </Application>
    <Application>
        <applicationId>292</applicationId>
        <cust_OnlineReferenceCheck>
            <Option>
                <id>197249</id>
            </Option>
        </cust_OnlineReferenceCheck>
        <cust_overallduedilligencestatus>
            <Option>
                <id>197276</id>
            </Option>
        </cust_overallduedilligencestatus>
    </Application>
    <Application>
        <applicationId>280</applicationId>
        <cust_AustralianWorkRights>
            <Option>
                <id>197250</id>
            </Option>
        </cust_AustralianWorkRights>
        <cust_overallduedilligencestatus>
            <Option>
                <id>197276</id>
            </Option>
        </cust_overallduedilligencestatus>
    </Application>
    <Application>
        <applicationId>292</applicationId>
        <cust_AustralianWorkRights>
            <Option>
                <id>197250</id>
            </Option>
        </cust_AustralianWorkRights>
        <cust_overallduedilligencestatus>
            <Option>
                <id>197276</id>
            </Option>
        </cust_overallduedilligencestatus>
    </Application>
    <Application>
        <applicationId>280</applicationId>
        <cust_NationalPoliceCheck>
            <Option>
                <id>197249</id>
            </Option>
        </cust_NationalPoliceCheck>
        <cust_overallduedilligencestatus>
            <Option>
                <id>197276</id>
            </Option>
        </cust_overallduedilligencestatus>
    </Application>
    <Application>
        <applicationId>292</applicationId>
        <cust_NationalPoliceCheck>
            <Option>
                <id>197249</id>
            </Option>
        </cust_NationalPoliceCheck>
        <cust_overallduedilligencestatus>
            <Option>
                <id>197276</id>
            </Option>
        </cust_overallduedilligencestatus>
    </Application>
</Applications>

需求：先按applicationId对<Application>节点分组，再对每组内的子元素按元素名去重，最终输出合并后的XML，预期结果如下：

<Applications>
    <Application>
        <applicationId>280</applicationId>
        <cust_IDCheckIDCollation>
            <Option>
                <id>197249</id>
            </Option>
        </cust_IDCheckIDCollation>
        <cust_overallduedilligencestatus>
            <Option>
                <id>197276</id>
            </Option>
        </cust_overallduedilligencestatus>
        <cust_OnlineReferenceCheck>
            <Option>
                <id>197249</id>
            </Option>
        </cust_OnlineReferenceCheck>
        <cust_AustralianWorkRights>
            <Option>
                <id>197250</id>
            </Option>
        </cust_AustralianWorkRights>
        <cust_NationalPoliceCheck>
            <Option>
                <id>197249</id>
            </Option>
        </cust_NationalPoliceCheck>
    </Application>
    <Application>
        <applicationId>292</applicationId>
        <cust_IDCheckIDCollation>
            <Option>
                <id>197249</id>
            </Option>
        </cust_IDCheckIDCollation>
        <cust_overallduedilligencestatus>
            <Option>
                <id>197276</id>
            </Option>
        </cust_overallduedilligencestatus>
        <cust_OnlineReferenceCheck>
            <Option>
                <id>197249</id>
            </Option>
        </cust_OnlineReferenceCheck>
        <cust_AustralianWorkRights>
            <Option>
                <id>197250</id>
            </Option>
        </cust_AustralianWorkRights>
        <cust_NationalPoliceCheck>
            <Option>
                <id>197249</id>
            </Option>
        </cust_NationalPoliceCheck>
    </Application>
</Applications>

原有XSLT 2.0代码的问题 #

编写的XSLT 2.0代码无法得到预期结果，核心问题是：

<xsl:element name="{current-grouping-key()}">
    <xsl:value-of select="(current-group())[1]"/>
</xsl:element>

<xsl:value-of>仅提取节点的文本内容，丢失了子元素结构（比如<Option>和<id>节点），导致输出中带嵌套结构的子节点无法完整保留。

XSLT 3.0的简便实现 #

基于原有分组逻辑，替换文本提取为完整节点复制，结合XSLT 3.0的语法特性，可实现简洁高效的解决方案：

<xsl:stylesheet version="3.0" 
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
    <xsl:output method="xml" version="1.0" encoding="UTF-8" indent="yes"/>
    <xsl:strip-space elements="*"/>

    <xsl:template match="/Applications">
        <xsl:copy>
            <!-- 按applicationId分组 -->
            <xsl:for-each-group select="Application" group-by="applicationId">
                <xsl:copy>
                    <!-- 对组内所有子元素按元素名去重 -->
                    <xsl:for-each-group select="current-group()/*" group-by="node-name()">
                        <!-- 复制分组后的第一个完整节点（包含所有子元素） -->
                        <xsl:copy-of select="current-group()[1]"/>
                    </xsl:for-each-group>
                </xsl:copy>
            </xsl:for-each-group>
        </xsl:copy>
    </xsl:template>
</xsl:stylesheet>

关键改进点 #

1. 用<xsl:copy-of>替代<xsl:value-of>，完整复制节点及其所有子元素、属性，保留原XML结构。
2. 使用node-name()代替name()，node-name()返回节点的QName，在有命名空间的场景下更严谨，无命名空间时效果一致。
3. XSLT 3.0原生支持分组逻辑，代码无需额外扩展即可高效运行。

内容的提问来源于stack exchange，提问作者Madhu N G