Question

我想使用模板来反转不同的XML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，这些章节已被撤销，但章节却没有。

我尝试使用两个模板来解决它，但它无法正常工作..

<?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

这个XSLT 2.0样式表：

<xsl:stylesheet version="2.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
    <xsl:strip-space elements="*"/>
    <xsl:template match="node()|@*" name="identity">
        <xsl:copy>
            <xsl:apply-templates select="node()|@*"/>
        </xsl:copy>
    </xsl:template>
    <xsl:template match="chapter|section">
        <xsl:copy>
            <xsl:apply-templates select="@*"/>
            <xsl:apply-templates>
                <xsl:sort select="tokenize(@title,'.')[last()]"
                          order="descending"
                          data-type="number"/>
            </xsl:apply-templates>
        </xsl:copy>
    </xsl:template>
</xsl:stylesheet>

输出：

<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>

Answer 2

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>

将此转换应用于以下XML文档：

<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>

生成了所需的正确结果（section元素的所有序列都已反转）：

<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>

逆转孩子的秩序

2 个答案: