如何使用 XSLT 从 XML 读取所有内部递归值

Question

需要从所有级别的递归中提取值XML structure.The所有级别的结构都相同

<regPackagingHierarchyList>
  <RegistrationPackagingHierarchy>
    <recordId>Level0</recordId>
    <regParentPackagingHierarchy>
      <RegistrationPackagingHierarchy>
        <recordId>level5</recordId>
        <regParentPackagingHierarchy>
          <RegistrationPackagingHierarchy>
            <recordId>level4</recordId>
            <regParentPackagingHierarchy>
              <RegistrationPackagingHierarchy>
                <recordId>level3</recordId>
                <regParentPackagingHierarchy>
                  <RegistrationPackagingHierarchy>
                    <recordId>level2</recordId>
                    <regParentPackagingHierarchy>
                      <RegistrationPackagingHierarchy>
                        <recordId>level1</recordId>
                      </RegistrationPackagingHierarchy>
                    </regParentPackagingHierarchy>
                  </RegistrationPackagingHierarchy>
                </regParentPackagingHierarchy>
              </RegistrationPackagingHierarchy>
            </regParentPackagingHierarchy>
          </RegistrationPackagingHierarchy>
        </regParentPackagingHierarchy>
      </RegistrationPackagingHierarchy>
    </regParentPackagingHierarchy>
  </RegistrationPackagingHierarchy>
</regPackagingHierarchyList>

Expected Result should be in the follwing format Level1,Level2,Level3,level4,Level5

Answer 1

1.0 中的一种方式可以是：

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">

<xsl:output method="xml" />

<xsl:template match="/">
    <xsl:for-each select="//regParentPackagingHierarchy//recordId">
        <xsl:sort select="position()" order="descending" />
        <xsl:value-of select="." />
        <xsl:if test="position() != last()">
            <xsl:text>,</xsl:text>
        </xsl:if>
    </xsl:for-each>
</xsl:template>
</xsl:stylesheet>

http://xsltfiddle.liberty-development.net/pPzifpv

编辑：

<xsl:value-of select="concat('Count(regPackagingHierarchyList) = ', count(//regPackagingHierarchyList))"/>
<xsl:value-of select="concat('Count(RegistrationPackagingHierarchy) = ', count(//RegistrationPackagingHierarchy))"/>

编辑 2：

http://xsltfiddle.liberty-development.net/pPzifpv/8

Answer 2

<xsl:template match="*">
    <xsl:apply-templates/>
</xsl:template>

<xsl:template match="recordId">
    <xsl:apply-templates/> 
    <!-- Apply all the templates beneath first and output your id after they got applied. -->
    <xsl:if test="ancestor::regParentPackagingHierarchy">
        <xsl:value-of select="."/>
    </xsl:if>
</xsl:template>

如何使用 XSLT 从 XML 读取所有内部递归值

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