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RE: Getting the XPath of a node
- From: "Macaulay,Malcolm (US)" <Malcolm dot Macaulay2 at cnare dot com>
- To: <xsl-list at lists dot mulberrytech dot com>
- Date: Wed, 4 Sep 2002 14:16:32 -0500
- Subject: RE: [xsl] Getting the XPath of a node
- Reply-to: xsl-list at lists dot mulberrytech dot com
Hi Dennis
I know you've have had plenty of replies on this one, but I just have to send you my 'generateXPath' named template. It work well and is in production on my current project (not wanting to suggest that any of the other replies would not work). Also I like it :-)
<xsl:template name="generateXPath">
<xsl:for-each select="ancestor::*">/<xsl:value-of select="name()"/>[<xsl:number/>]</xsl:for-each>/<xsl:value-of select="name()"/>[<xsl:number/>]</xsl:template>
Call this template at the context where you want the unique XPath inserted into the output.
Note: keep this template all on one line (i.e. no whitespace or nice indenting) - I use the generated XPath in a javascript function - whitespace in the template = gaps in the generate XPath = stuffed up the javascript function.
Hope that helps.
cheers
Malcolm
-----Original Message-----
From: Dennis [mailto:maillistboxes@yahoo.com]
Sent: Wednesday, September 04, 2002 4:03 AM
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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