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RE: Getting the XPath of a node
- From: "Michael Kay" <michael dot h dot kay at ntlworld dot com>
- To: <xsl-list at lists dot mulberrytech dot com>
- Date: Wed, 4 Sep 2002 12:27:58 +0100
- Subject: RE: [xsl] Getting the XPath of a node
- Reply-to: xsl-list at lists dot mulberrytech dot com
<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@ntlworld.com
work: Michael.Kay@softwareag.com
> -----Original Message-----
> From: owner-xsl-list@lists.mulberrytech.com
> [mailto:owner-xsl-list@lists.mulberrytech.com] On Behalf Of Dennis
> Sent: 04 September 2002 10:03
> To: xsl-list@lists.mulberrytech.com
> 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@netscape.com</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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