如何在XSLT中显示最后两个节点

您能帮助我在XSLT中获得最后两个节点。http://xsltransform.net/bwdwsj/1

ExpectOutput

<h1>Preview your result as PDF when doctype is set to XML and your document starts with
      root element of XSL-FO. Apache FOP is used to generate the PDF
   </h1>
   <h1>Added some links to useful XSLT sites</h1>

XSLT代码

 <xsl:template match="/">
      <hmtl>
        <head>
          <title>New Version!</title>
        </head>
        <xsl:for-each select="ul/li[last() &gt;2]">
            <h1><xsl:value-of select="."/></h1>
        </xsl:for-each>
      </hmtl>
    </xsl:template>

XML

<?xml version="1.0" encoding="UTF-8"?>
    <ul>
        <li>A new XSLT engine is added: Saxon 9.5 EE, with a license (thank you Michael Kay!)</li>
        <li>XSLT 3.0 support when using the new Saxon 9.5 EE engine!</li>
        <li>Preview your result as HTML when doctype is set to HTML (see this example)</li>
        <li>Preview your result as PDF when doctype is set to XML and your document starts with root element of XSL-FO. Apache FOP is used to generate the PDF</li>
        <li>Added some links to useful XSLT sites</li>
    </ul>

简短但有效： ul/li[position() >= last()-1]

当前处理节点的位置必须大于或等于最后一个元素减1的索引。

显示：http：//xsltransform.net/bwdwsj/2

您可以在XPath表达式下使用：

/ul/li[position()>count(/ul/li)-2]

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