How do you output the current element path in XSLT?

Question

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.

Answer 1

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:template>

  <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>

</xsl:stylesheet>

XML Output

<Doc path="/Doc[1]">
   <Ele1 path="/Doc[1]/Ele1[1]">
      <Ele11 path="/Doc[1]/Ele1[1]/Ele11[1]">
         <Ele111 path="/Doc[1]/Ele1[1]/Ele11[1]/Ele111[1]">
            <foo path="/Doc[1]/Ele1[1]/Ele11[1]/Ele111[1]/foo[1]"/>
            <foo path="/Doc[1]/Ele1[1]/Ele11[1]/Ele111[1]/foo[2]"/>
            <bar path="/Doc[1]/Ele1[1]/Ele11[1]/Ele111[1]/bar[1]"/>
            <foo path="/Doc[1]/Ele1[1]/Ele11[1]/Ele111[1]/foo[3]"/>
            <foo path="/Doc[1]/Ele1[1]/Ele11[1]/Ele111[1]/foo[4]"/>
            <bar path="/Doc[1]/Ele1[1]/Ele11[1]/Ele111[1]/bar[2]"/>
            <bar path="/Doc[1]/Ele1[1]/Ele11[1]/Ele111[1]/bar[3]"/>
         </Ele111>
      </Ele11>
   </Ele1>
   <Ele2 path="/Doc[1]/Ele2[1]"/>
</Doc>

Here's another version that only outputs the positional predicate if it's needed. This example is also different in that it's just outputting the path instead of adding an attribute.

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

    <xsl:template match="text()"/>

    <xsl:template match="*">
        <xsl:for-each select="ancestor-or-self::*">
            <xsl:value-of select="concat('/',local-name())"/>
            <!--Predicate is only output when needed.-->
            <xsl:if test="(preceding-sibling::*|following-sibling::*)[local-name()=local-name(current())]">
                <xsl:value-of select="concat('[',count(preceding-sibling::*[local-name()=local-name(current())])+1,']')"/>
            </xsl:if>
        </xsl:for-each>
        <xsl:text>&#xA;</xsl:text>
        <xsl:apply-templates select="node()"/>
    </xsl:template>

</xsl:stylesheet>

using the input above, this stylesheet outputs:

/Doc
/Doc/Ele1
/Doc/Ele1/Ele11
/Doc/Ele1/Ele11/Ele111
/Doc/Ele1/Ele11/Ele111/foo[1]
/Doc/Ele1/Ele11/Ele111/foo[2]
/Doc/Ele1/Ele11/Ele111/bar[1]
/Doc/Ele1/Ele11/Ele111/foo[3]
/Doc/Ele1/Ele11/Ele111/foo[4]
/Doc/Ele1/Ele11/Ele111/bar[2]
/Doc/Ele1/Ele11/Ele111/bar[3]
/Doc/Ele2

Answer 2

Don't think this is built into XPath, you probably need a recursive template, like the one here, which I've based this example on. It will walk every element in an XML document and output the path to that element in a style similar to the one you've described.

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" 
      xmlns:xs="http://www.w3.org/2001/XMLSchema"
      exclude-result-prefixes="xs"
      version="2.0">

    <xsl:template match="/">
        <paths>
            <xsl:apply-templates/>
        </paths>
    </xsl:template>

    <xsl:template match="//*">
        <path>
        <xsl:for-each select="ancestor-or-self::*">
            <xsl:call-template name="print-step"/>
        </xsl:for-each>
        </path>
        <xsl:apply-templates select="*"/>
    </xsl:template>

    <xsl:template name="print-step">
        <xsl:text>/</xsl:text>
        <xsl:value-of select="name()"/>
        <xsl:text>[</xsl:text>
        <xsl:value-of select="1+count(preceding-sibling::*)"/>
        <xsl:text>]</xsl:text>
    </xsl:template>

</xsl:stylesheet>

There are a few complications; consider this tree:

<root>
  <child/>
  <child/>
</root>

How do you tell the difference between the two child nodes? So you need some index into your item-sequence, child1 and child[2], for example.

Answer 3

Since XPath 3.0 as supported by Saxon 9.8 (all editions) or Saxon 9.7 with version="3.0" in the XSLT and XmlPrime 4 (using --xt30) as well as 2017 releases of Altova (using version="3.0" stylesheets) there is the built-in path function (https://www.w3.org/TR/xpath-functions-30/#func-path, https://www.w3.org/TR/xpath-functions-31/#func-path) which for an input like

<?xml version="1.0" encoding="UTF-8"?>
<Doc>
    <Ele1>
        <Ele11>
            <Ele111>
                <foo/>
                <foo/>
                <bar/>
                <foo/>
                <foo/>
                <bar/>
                <bar/>
            </Ele111>
        </Ele11>
    </Ele1>
    <Ele2/>  
</Doc>

and a stylesheet like

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
    xmlns:xs="http://www.w3.org/2001/XMLSchema"
    xmlns:math="http://www.w3.org/2005/xpath-functions/math"
    exclude-result-prefixes="xs math"
    version="3.0">

    <xsl:output method="text"/>

    <xsl:template match="/">
        <xsl:value-of select="//*/path()" separator="&#10;"/>
    </xsl:template>

</xsl:stylesheet>

gives the output

/Q{}Doc[1]
/Q{}Doc[1]/Q{}Ele1[1]
/Q{}Doc[1]/Q{}Ele1[1]/Q{}Ele11[1]
/Q{}Doc[1]/Q{}Ele1[1]/Q{}Ele11[1]/Q{}Ele111[1]
/Q{}Doc[1]/Q{}Ele1[1]/Q{}Ele11[1]/Q{}Ele111[1]/Q{}foo[1]
/Q{}Doc[1]/Q{}Ele1[1]/Q{}Ele11[1]/Q{}Ele111[1]/Q{}foo[2]
/Q{}Doc[1]/Q{}Ele1[1]/Q{}Ele11[1]/Q{}Ele111[1]/Q{}bar[1]
/Q{}Doc[1]/Q{}Ele1[1]/Q{}Ele11[1]/Q{}Ele111[1]/Q{}foo[3]
/Q{}Doc[1]/Q{}Ele1[1]/Q{}Ele11[1]/Q{}Ele111[1]/Q{}foo[4]
/Q{}Doc[1]/Q{}Ele1[1]/Q{}Ele11[1]/Q{}Ele111[1]/Q{}bar[2]
/Q{}Doc[1]/Q{}Ele1[1]/Q{}Ele11[1]/Q{}Ele111[1]/Q{}bar[3]
/Q{}Doc[1]/Q{}Ele2[1]

That output is not as compact in case of lack of namespaces as most hand-made attempts but the format has the advantage (at least given XPath 3.0 or 3.1 support) to allow for namespaces being used and get a format for the returned path that does not require the user of the path expression to set up any namespace bindings to evaluate it.

Answer 4

I'm not sure which XSLT processor you're using, but if it is Saxon, you can use extension function path(). Other processors may have same functionality.

Answer 5

You can use the ancestor XPath Axes to walk all parent, and grandparents.

<xsl:for-each select="ancestor::*">...