RE: [xsl] Getting the XPath of a node

["Michael Kay" <michael.h.kay@xxxxxxxxxxxx>]

<xsl:for-each select="ancestor-or-self::*">
<xsl:value-of select="concat(
   '/',
   name(.),
   '[',
   count(preceding-sibling::*[name(.)=name(current())])-1,
   ']')"/>
</xsl:for-each>

(Saxon has a saxon:path extension function)

Michael Kay
Software AG
home: Michael.H.Kay@xxxxxxxxxxxx
work: Michael.Kay@xxxxxxxxxxxxxx 
   

> -----Original Message-----
> From: owner-xsl-list@xxxxxxxxxxxxxxxxxxxxxx 
> [mailto:owner-xsl-list@xxxxxxxxxxxxxxxxxxxxxx] On Behalf Of Dennis
> Sent: 04 September 2002 10:03
> To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
> Subject: [xsl] Getting the XPath of a node
> 
> 
> Hi All,
> 
> Is there any way to get the XPath of a particular
> element and attribute in match template???
> 
> Say if I have following XML:
> <Person id="12345">
>   <Name>Dennis</Name>
>   <Company>Netscape</Company>
>   <Address>Mountain View</Address>
>   <Email>dennis@xxxxxxxxxxxx</Email>
> </Person>
> 
> ----The XSL to print XPath---
> <xsl:template match="Company">
> //Print the XPath of Company as /Person/Company
> </xsl:template>
> More templates corresponding to each element.
> 
> How do I do this...any thoughts???
> 
> Thanks
> Dennis
> 
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