如何在 xslt 中显示最后两个节点？答案

【问题标题】：how to show last two node in xslt?如何在 xslt 中显示最后两个节点？
【发布时间】：2017-07-09 08:44:56
【问题描述】：

您能帮我获取 xslt 中的最后两个节点吗？这是我的代码 http://xsltransform.net/bwdwsJ/1

期望输出

<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>

【问题讨论】：

标签： jquery xml xslt xpath xslt-1.0

【解决方案1】：

你可以使用下面的XPath表达式：

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

【讨论】：

【解决方案2】：

简短但有效：ul/li[position() &gt;= last()-1]

当前处理节点的位置必须大于或等于最后一个元素的索引减1。

显示在：http://xsltransform.net/bwdwsJ/2

【讨论】：