如何在XSLT中删除XML重复项并设置连续序列值（10、20…）

现有XML文件存在重复的resourceMemberList节点，已编写XSLT实现去重逻辑，但需要将去重后的sequence节点设置为10、20这类连续递增的序列值，添加序列设置逻辑时会破坏原有去重功能，寻求可行解决方案。

原XML代码 #

<?xml version="1.0" encoding="UTF-8"?>

<serviceInvocationRequests>
  <serviceInvocationRequest>
    <requests>
      <workCenter>04901</workCenter>
      <description>Some random description</description>
    <resourceMemberList>
        <resource>R4902A01</resource>
        <sequence>10</sequence>
      </resourceMemberList>
            <resourceMemberList>
        <resource>R4902A02</resource>
        <sequence>10</sequence>
      </resourceMemberList>
            <resourceMemberList>
        <resource>R4902A01</resource>
        <sequence>10</sequence>
      </resourceMemberList>
            <resourceMemberList>
        <resource>R4902A02</resource>
        <sequence>10</sequence>
      </resourceMemberList>
        </requests>
  </serviceInvocationRequest>
</serviceInvocationRequests>

现有去重XSLT代码 #

<?xml version="1.0" encoding="UTF-8"?>

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

    <xsl:key match="resourceMemberList" name="group-key" use="concat(resource, '|')"/>

    <xsl:template match="node() | @*">
        <xsl:copy>
            <xsl:apply-templates select="node() | @*"/>
        </xsl:copy>
    </xsl:template>

    <xsl:template match="resourceMemberList[generate-id() = generate-id(key('group-key', concat(resource, '|'))[1])]">
        <xsl:copy>
            <xsl:apply-templates/>
        </xsl:copy>
    </xsl:template>

    <xsl:template match="resourceMemberList"/>

</xsl:stylesheet>

期望输出XML #

<?xml version="1.0" encoding="UTF-8"?>
<serviceInvocationRequests>
   <serviceInvocationRequest>
      <requests>
         <workCenter>04901</workCenter>
         <description>Some random description</description>
         <resourceMemberList>
            <resource>R4902A01</resource>
            <sequence>10</sequence>
         </resourceMemberList>
         <resourceMemberList>
            <resource>R4902A02</resource>
            <sequence>20</sequence>
         </resourceMemberList>
      </requests>
   </serviceInvocationRequest>
</serviceInvocationRequests>

解决方案 #

修改后的XSLT将保留Muenchian分组去重逻辑，同时为每个唯一的resourceMemberList生成递增的sequence值：

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

    <xsl:key match="resourceMemberList" name="group-key" use="resource"/>

    <!-- 身份模板：复制所有未被特殊匹配的节点和属性 -->
    <xsl:template match="node() | @*">
        <xsl:copy>
            <xsl:apply-templates select="node() | @*"/>
        </xsl:copy>
    </xsl:template>

    <!-- 匹配requests节点，遍历所有唯一的resourceMemberList -->
    <xsl:template match="requests">
        <xsl:copy>
            <xsl:apply-templates select="node()[not(self::resourceMemberList)]"/>
            <!-- 按顺序处理每个唯一的resourceMemberList -->
            <xsl:for-each select="resourceMemberList[generate-id() = generate-id(key('group-key', resource)[1])]">
                <xsl:copy>
                    <xsl:apply-templates select="resource"/>
                    <!-- 生成递增的sequence值：位置*10 -->
                    <sequence>
                        <xsl:value-of select="position() * 10"/>
                    </sequence>
                </xsl:copy>
            </xsl:for-each>
        </xsl:copy>
    </xsl:template>

    <!-- 忽略所有原始的resourceMemberList节点，避免重复输出 -->
    <xsl:template match="resourceMemberList"/>

</xsl:stylesheet>

逻辑说明 #

1. 保留去重逻辑：依然使用Muenchian分组，通过key按resource值对resourceMemberList分组，确保每个resource只保留一个实例。
2. 重新生成sequence：在requests节点的模板中，先复制除resourceMemberList外的所有子节点，再遍历所有唯一的resourceMemberList，用position() * 10生成10、20、30...的递增序列值。
3. 避免冲突：单独处理requests节点下的resourceMemberList输出，同时忽略原始的resourceMemberList节点，确保去重和序列生成逻辑互不干扰。

内容的提问来源于stack exchange，提问作者Maksim Sirotkin