[xsl] xsl transform

Subject: [xsl] xsl transform
From: "Philippe LAPLANCHE" <philippe.laplanche@xxxxxxxxxxxx>
Date: Tue, 28 Jun 2005 15:58:19 +0200

Does somebody has a neet way to convert this tree :
<?xml version="1.0" encoding="UTF-8"?>

Into this one :
<?xml version="1.0" encoding="UTF-8"?>
<root fields="param1|param2|param3 ... ">
	<e a="p1" b="p2" c="p3" d="p4" e="p5" f="p6" ... z="p26" aa="p27" ab="p28"
... az="p52" ba="p53" .... zz="p702" />

I'm thinking of using a list of attributes names but I can't find how-to

First I don't know how to define a list, I want something like that:

<xsl:variable name="attributes">
In order to access attributes[n] and use it as an xsl:attribute's name

I'd have templates that look like that :

<xsl:template match="/results"> (this works fine)
		<xsl:attribute name="fields">
			<xsl:for-each select="row[1]/*">
				<xsl:value-of select="local-name()"/>
				<xsl:if test="not(position()=last())">|</xsl:if>

<xsl:template match="row"> (this I don't know how to make it work)
		<xsl:for-each select="*">
			<xsl:attribute name="????">
				<xsl:value-of select="."/>

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