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Re: XPath position in XSLT

• To: "Schaik L.B. van" <L_B_van_Schaik@L...>
• Subject: Re: XPath position in XSLT
• From: "m batsis" <mbatsis@n...>
• Date: Tue, 04 Feb 2003 18:13:03 +0200
• Cc: "'xml-dev@l...'" <xml-dev@l...>
• In-reply-to: <9A11FBB85B97D311A6C700508B5A188C0100E2B1@u...>
• Organization: Netsmart S.A., Greece
• References: <9A11FBB85B97D311A6C700508B5A188C0100E2B1@u...>
• Reply-to: mbatsis@n...
• User-agent: Mozilla/5.0 (Windows; U; Windows NT 5.0; en-US; rv:1.2.1) Gecko/20021130

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

• References:
  • XPath position in XSLT
    • From: "Schaik, L.B. van" <L_B_van_Schaik@L...>

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