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   Re: [xml-dev] XPath position in XSLT

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I wouldn't give a try on this but I wanted it for myself anyway so... 
Please note that there are more appropriate forums to ask XSLT related 
questions:

http://www.mulberrytech.com/xsl/xsl-list/
http://www.topxml.com/xsltalk/default.asp


Anyway, it took less than 5 minutes to do this, so I'm sure there are 
better ways but here it goes:

<?xml version="1.0" encoding="UTF-8"?>
<xsl:transform version="1.0" 
xmlns:xsl="http://www.w3.org/1999/XSL/Transform";>
	<xsl:output method="xml" version="1.0" encoding="UTF-8"/>

	<xsl:template match="text()">
	    <text><xsl:attribute name="position">//<xsl:for-each 
select="ancestor::*[ancestor::*]"><xsl:variable name="nm" 
select="name()"/><xsl:value-of select="$nm"/>[<xsl:value-of 
select="count(preceding-sibling::*[name() = 
$nm])+1"/>]/</xsl:for-each></xsl:attribute><xsl:value-of select="."/>
	    </text>
	</xsl:template>

</xsl:transform >


Use at your own risk ;-)

Cheers,

Manos





 

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