How to Get XPath to an XML Node

There are several XSL templates floating around that get a human readable path to a node. However, I found none that cover the case when there are multiple nodes of the same element at the same level. I came up with a complicated solution that kept the output human readable, but the solution for something machine readable is actually simple.

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
  <xsl:output indent="yes"/>
  <xsl:template match="@*|node()">
      <xsl:copy>        
        <xsl:attribute name="position"><xsl:call-template name="xpath.position"></xsl:call-template</xsl:attribute>
        <xsl:apply-templates select="@*|node()"/>
      </xsl:copy>
  </xsl:template>

  <xsl:template name="xpath.position">
    <xsl:param name="node" select="."></xsl:param> <!-- default value is current node -->
    <xsl:param name="path" select="''"></xsl:param> <!-- default value is empty string -->

    <!-- prepend the current position to $path, save it as next.path -->
    <xsl:variable name="next.path"><xsl:text>*[position()=</xsl:text><xsl:value-of select="$node/count(preceding-sibling::*)+1"/><xsl:text>]</xsl:text>
      <xsl:if test="$path != ''">/</xsl:if>
      <xsl:value-of select="$path"></xsl:value-of>
    </xsl:variable>

    <xsl:choose>
      <xsl:when test="$node/parent::*"> <!-- when there is a parent node -->
        <xsl:call-template name="xpath.position">
          <xsl:with-param name="node" select="$node/parent::*"></xsl:with-param>
          <xsl:with-param name="path" select="$next.path"></xsl:with-param>
        </xsl:call-template>
      </xsl:when>
      <xsl:otherwise>
        <xsl:text>/</xsl:text>
        <xsl:value-of select="$next.path"></xsl:value-of>
      </xsl:otherwise>
    </xsl:choose>
  </xsl:template>
  
</xsl:stylesheet>