Question

我的任务是打破祖先<ol>的嵌套<ol>并使每个<ol>元素处于同一级别。我有这个xml，

<main>
  <ol>
    <li>The above</li>
    <li>Tprojects.</li>
    <li>FreeSpan sections.</li>
    <li>The above
        <ol>
            <li>Maximum
                <ol>
                    <li>Middle</li>
                </ol>
            </li>
            <li>Ultimate</li>
        </ol>
    </li>
    <li>The above indicative</li>
    <li>Appropriate Live</li>
    <li>The above Indicative</li>
  </ol>
</main>

所以，预期输出，

<main>
    <ol>
        <li>The above</li>
        <li>Tprojects.</li>
        <li>FreeSpan sections.</li>
        <li>The above</li>
    </ol>
    <ol>
        <li>Maximum</li>
    </ol>
    <ol>
        <li>Middle</li>
    </ol>
    <ol>
        <li>Ultimate</li>
    </ol>
    <ol>
        <li>The above indicative</li>
        <li>Appropriate Live</li>
        <li>The above Indicative</li>
    </ol>
</main>

我尝试使用for-each实现此功能，但无法正确完成，这是我尝试的方法，

<xsl:template match="ol[descendant::ol]">
        <xsl:for-each select="li">
            <ol>
                <li>
                    <xsl:apply-templates select="node()[not(self::ol)]"/>
                </li>
            </ol>
            <xsl:apply-templates select="ol"/>
        </xsl:for-each>
    </xsl:template>

Answer 1

对于您的两个样本，我都会得到正确的输出

<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
    xmlns:xs="http://www.w3.org/2001/XMLSchema"
    exclude-result-prefixes="#all"
    version="3.0">

  <xsl:mode on-no-match="shallow-copy"/>

  <xsl:output indent="yes"/>

  <xsl:template match="ol[descendant::ol]">
      <xsl:for-each-group select="descendant::li" group-starting-with="li[. is ../li[1]]">
          <xsl:for-each-group select="current-group()" group-ending-with="li[. is ../li[last()]]">
              <ol>
                  <xsl:apply-templates select="current-group()"/>
              </ol>              
          </xsl:for-each-group>
      </xsl:for-each-group>
  </xsl:template>

  <xsl:template match="li">
      <xsl:copy>
          <xsl:apply-templates select="text()/normalize-space()"/>
      </xsl:copy>
  </xsl:template>

</xsl:stylesheet>

https://xsltfiddle.liberty-development.net/94rmq79/和https://xsltfiddle.liberty-development.net/94rmq79/1

它是XSLT 3，但是对于XSLT 2处理器，您当然可以替换由显式模板用于身份转换的xsl:mode声明。

XSLT-中断父元素

1 个答案: