Re: [xsl] Write context xpath expression

[Syd Bauman <Syd_Bauman@xxxxxxxxx>]

I have successfully used something like the following:
---------
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="1.0"
     xmlns:xsl="http://www.w3.org/1999/XSL/Transform";>
  <xsl:template match="/">
    <xsl:apply-templates select=".//c"/>
  </xsl:template>
  <xsl:template match="c">
    <xsl:message>
      <xsl:for-each select="ancestor-or-self::*">
        <xsl:value-of select="concat( '/', local-name(.) )"/>
      </xsl:for-each>
      <xsl:text>[</xsl:text>
      <xsl:value-of select="position()"/>
      <xsl:text>]</xsl:text>
    </xsl:message>
  </xsl:template>
</xsl:stylesheet>
---------
Of course this drops the namespace information. (Which could be
important in your situation, but didn't matter in mine where there
were no name collisions in the documents). Also, I think this may
give you undesired counts in the square brackets if you use a
predicate to select only some of the <c> elements. But if you only
want the path, that for-each should do the job, I think.

Note that this sort of question is often easily answered on Dave
Pawson's FAQ pages. See
http://www.dpawson.co.uk/xsl/sect2/N6052.html#d9122e58. 

> In XSL 1.0, how can I write out the XPATH expression for the
> current context? Example:
> <a>
>   <b>
>     <c />
>     <c /> <!-- assume this is the context -->
>     <c />
>   </b>
> </a>
> Assuming <c/> is the current context I would like to determine the
> xpath pattern like this: /a/b/c
> How is this possible? I get how I could walk parent;;node() but I'd
> be writing the pattern backwards! Any ideas?