XSL:获取不带exslt的可变数据:node-set

Question

在PHP中使用本机XSL库.是否有可能在变量中获取节点值而无需每次都通过exslt:node-set调用它......它很长而且很难看.

<xsl:variable name="mydata">
    <foo>1</foo>
    <bar>2</bar>
</xsl:variable>


<!-- How currently being done -->
<xsl:value-of select="exslt:node-set($mydata)/foo" />


<!-- I want to be able to do this -->
<xsl:value-of select="$mydata/foo" />

Answer 1

<xsl:variable name="mydata">
 <foo>1</foo>
 <bar>2</bar>
</xsl:variable>

<!-- How currently being done -->
<xsl:value-of select="exslt:node-set($mydata)/foo" />

<!-- I want to be able to do this -->
<xsl:value-of select="$mydata/foo" />

如果变量的内容被静态定义的,那么它是可以从XPath表达式访问它不使用的xxx:node-set()扩展功能.

示例:

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

 <xsl:variable name="mydata">
  <foo>1</foo>
  <bar>2</bar>
 </xsl:variable>

 <xsl:template match="/">
  <xsl:value-of select=
   "document('')/*/xsl:variable[@name='mydata']/bar"/>
 </xsl:template>
</xsl:stylesheet>

当此转换应用于任何XML文档(未使用)时,将生成所需的正确结果:

2