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XSL Copy
- To: <xsl-list at mulberrytech dot com>
- Subject: XSL Copy
- From: "Carlos Sanchez" <carlos at ktsi dot com>
- Date: Tue, 11 Apr 2000 11:18:44 -0500
- Reply-To: xsl-list at mulberrytech dot com
Hi,
I have the following XML
<?xml version="1.0" ?>
<results>
<timeSeriesDetail>
<timeSeries>
<timeSeriesType>Equity</timeSeriesType>
<timeSeriesKey>DAX</timeSeriesKey>
</timeSeries>
<obs>
<date>19990817</date>
<level>5324.02</level>
<return>0</return>
</obs>
<timeSeriesDetail>
<timeSeriesDetail>
<timeSeries>
<timeSeriesType>Equity</timeSeriesType>
<timeSeriesKey>CAC40</timeSeriesKey>
</timeSeries>
<obs>
<date>19990817</date>
<level>124.02</level>
<return>2</return>
</obs>
<timeSeriesDetail>
<results>
I want to be able to get the timeSeriesDetail node for DAX. I am using the
following XSL
<?xml version="1.0"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
version="1.0">
<xsl:param name="tsType" select="Equity"/>
<xsl:param name="tsName" select="DAX"/>
<xsl:template match="//timeSeriesDetail/timeSeries[timeSeriesType=$tsType
and timeSeriesKey=$tsName]">
<xsl:copy>
<xsl:apply-templates select="*|@*|text()"/>
</xsl:copy>
</xsl:template>
</xsl:stylesheet>
However, when I run this the tag names are stripped off from the output and
I only get the node contents. What am I doing wrong?
Thx,
Carlos Sanchez
RiskMetrics
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