xslt - 颠倒 child 的顺序

Question

我想使用模板来反转 XML 的不同序列；
例如 :

<book title="Definitive XML Schema">
  <author first="Priscilla" />
  <chapter title="[I] ">
    <section title="[I.1]" />
    <section title="[I.2]">
      <section title="[I.2.1]" />
      <section title="[I.2.2]" />
    </section>
    <section title="[I.3] ">
      <section title="[I.3.1]" />
    </section>
  </chapter>
  <chapter title="[II]">
    <section title="[II.1]" />
    <section title="[II.2]">
      <section title="[II.2.1]" />
      <section title="[II.2.2]" />
    </section>
  </chapter>
</book>

我想得到这样的输出:这是我的 xsl。

<?xml version="1.0" encoding="UTF-8"?>
<book title="Definitive XML Schema">
   <author first="Priscilla"/>
   <chapter title="[I]">
      <section title="[I.3]">
         <section title="[I.3.1]"/>
      </section>
      <section title="[I.2]">
         <section title="[I.2.2]"/>
         <section title="[I.2.1]"/>
      </section>
      <section title="[I.1]"/>
   </chapter>
   <chapter title="[II]">
      <section title="[II.2]">
         <section title="[II.2.2]"/>
         <section title="[II.2.1]"/>
      </section>
      <section title="[II.1]"/>
   </chapter>
</book>

是的，章节颠倒了，但章节没有颠倒。

我尝试使用两个模板来解决它，但它无法工作..

<?xml version="1.0" encoding="UTF-8"?>
<xsl:transform version="2.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:xs="http://www.w3.org/2001/XMLSchema" exclude-result-prefixes="xs">
<xsl:output method="xml"  version="1.0" encoding="UTF-8" indent ="yes"/>
 <xsl:template match="/">
 <xsl:apply-templates/>
 <xsl:text>&#10;</xsl:text>
 </xsl:template>

<xsl:template match="book">
  <xsl:copy>
  <xsl:sequence select="@title"/>
  <xsl:sequence select="author"/>
  <xsl:apply-templates select="chapter">
    <xsl:with-param name="seq" select="section"/>

     </xsl:apply-templates>
    </xsl:copy>
  </xsl:template>

 <xsl:template match ="chapter|section" as="element()">
  <xsl:param name="seq" as="element(section)*"/>
   <xsl:copy>
     <xsl:sequence select="@title"/>
     <xsl:if test="not(empty($seq))">
    <xsl:apply-templates select="chapter">
        <xsl:with-param name="seq" select="$seq[position()>1]"/>
    </xsl:apply-templates> 
     <xsl:apply-templates select="$seq[1]"/>    
    </xsl:if>
  </xsl:copy>
 </xsl:template>
 </xsl:transform>

Answer 1

这是一个更简单的 XSLT 1.0(和 2.0)解决方案 :

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

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

 <xsl:template match="*[section]">
  <xsl:copy>
   <xsl:copy-of select="@*"/>
   <xsl:apply-templates>
     <xsl:sort select="position()"
      data-type="number" order="descending"/>
   </xsl:apply-templates>
  </xsl:copy>
 </xsl:template>
</xsl:stylesheet>

当此转换应用于以下 XML 文档时 :

<book title="Definitive XML Schema">
  <author first="Priscilla" />
  <chapter title="[I] ">
    <section title="[I.1]" />
    <section title="[I.2]">
      <section title="[I.2.1]" />
      <section title="[I.2.2]" />
    </section>
    <section title="[I.3] ">
      <section title="[I.3.1]" />
    </section>
  </chapter>
  <chapter title="[II]">
    <section title="[II.1]" />
    <section title="[II.2]">
      <section title="[II.2.1]" />
      <section title="[II.2.2]" />
    </section>
  </chapter>
</book>

产生想要的、正确的结果(section 元素的所有序列都颠倒了) :

<book title="Definitive XML Schema">
   <author first="Priscilla"/>
   <chapter title="[I] ">
      <section title="[I.3] ">
         <section title="[I.3.1]"/>
      </section>
      <section title="[I.2]">
         <section title="[I.2.2]"/>
         <section title="[I.2.1]"/>
      </section>
      <section title="[I.1]"/>
   </chapter>
   <chapter title="[II]">
      <section title="[II.2]">
         <section title="[II.2.2]"/>
         <section title="[II.2.1]"/>
      </section>
      <section title="[II.1]"/>
   </chapter>
</book>