Altova Mailing List Archives>Archive Index >microsoft.public.xsl Archive Home >Recent entries >Thread Prev - Re: Transforming enumerated list elements [Thread Next] Re: Transforming enumerated list elementsTo: 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 < $max">
<xsl:variable name="quantity" select="concat('Quantity',
format-number($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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