How do you output the current element path in XSLT?

In XSLT, is there a way to determine where you are in an XML document when processing an element?

Example: Given the following XML Doc Fragment...

<Doc>
  <Ele1>
    <Ele11>
      <Ele111>
      </Ele111>
    </Ele11>
  </Ele1>
  <Ele2>
  </Ele2>
</Doc>

In XSLT, if my context is the Element "Ele111", how can I get XSLT to output the full path? I would want it to output: "/Doc/Ele1/Ele11/Ele111".

The context of this question: I have a very large, very deep document that I want to traverse exhaustively (generically using recursion), and if I find an element with a particular attribute, I want to know where I found it. I suppose I could carry along my current path as I traverse, but I would think XSLT/XPath should know.

xml xslt xpath

Edit Report

The currently accepted answer will return incorrect paths. For example, the element Ele2 in the OP sample XML would return the path /Doc[1]/Ele2[2]. It should be /Doc[1]/Ele2[1].

Here's a similar XSLT 1.0 template that returns the correct paths:

  <xsl:template name="genPath">
    <xsl:param name="prevPath"/>
    <xsl:variable name="currPath" select="concat('/',name(),'[',
      count(preceding-sibling::*[name() = name(current())])+1,']',$prevPath)"/>
    <xsl:for-each select="parent::*">
      <xsl:call-template name="genPath">
        <xsl:with-param name="prevPath" select="$currPath"/>
      </xsl:call-template>
    </xsl:for-each>
    <xsl:if test="not(parent::*)">
      <xsl:value-of select="$currPath"/>      
    </xsl:if>
  </xsl:template>

Here's an example that will add a path attribute to all elements.

XML Input

<Doc>
  <Ele1>
    <Ele11>
      <Ele111>
        <foo/>
        <foo/>
        <bar/>
        <foo/>
        <foo/>
        <bar/>
        <bar/>
      </Ele111>
    </Ele11>
  </Ele1>
  <Ele2/>  
</Doc>

XSLT 1.0

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
  <xsl:output indent="yes"/>
  <xsl:strip-space elements="*"/>

  <xsl:template match="text()|@*">
    <xsl:copy>
      <xsl:apply-templates select="node()|@*"/>
    </xsl:copy>
  </xsl:template>

  <xsl:template match="*">
    <xsl:copy>
      <xsl:attribute name="path">
        <xsl:call-template name="genPath"/>
      </xsl:attribute>
      <xsl:apply-templates select="node()|@*"/>
    </xsl:copy>    
  </xsl:templat


                        
                        
                            answered 2012-04-11T19:25:57.883

How do you output the current element path in XSLT?

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