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Re: Transforming enumerated list elements

From: "dansherpa@-----.---" <---------@-----.--->
To: NULL
Date: 11/2/2005 11:43:00 AM
Thanks Kryptomoon.  I've tried this and still get it to work.  This is
my current XSL:

<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
	<xsl:output method="xml" version="1.0" encoding="UTF-8" indent="yes"/>
	<xsl:template match="LineItems">
		<LineItems>
			<xsl:call-template name="loopMe">
				<xsl:with-param name="counter" select="0"/>
				<xsl:with-param name="max" select="count(Quantity/*)"/>
			</xsl:call-template>
		</LineItems>
	</xsl:template>
	<xsl:template name="loopMe">
		<xsl:param name="counter"/>
		<xsl:param name="max"/>
		<xsl:if test="$counter &lt; $max">
			<xsl:variable name="quantity" select="concat('Quantity',
format-number($counter, '0000'))"/>
			<xsl:variable name="description" select="concat('Description',
format-number($counter, '0000'))"/>
			<LineItem>
				<Quantity>
					<xsl:value-of select="//*[name()='$quantity']"/>
				</Quantity>
				<Description>
					<xsl:value-of select="//*[name()='$description']"/>
				</Description>
			</LineItem>
			<xsl:call-template name="loopMe">
				<xsl:with-param name="counter" select="$counter + 1"/>
				<xsl:with-param name="max" select="$max"/>
			</xsl:call-template>
		</xsl:if>
	</xsl:template>
</xsl:stylesheet>


The output is formatted correctly but does not contain any of the
values:

<?xml version="1.0" encoding="UTF-8"?>
<LineItems>
	<LineItem>
		<Quantity></Quantity>
		<Description></Description>
	</LineItem>
	<LineItem>
		<Quantity></Quantity>
		<Description></Description>
	</LineItem>
</LineItems>

Anyone have any thoughts?  Thanks.



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