RE: XSL Copy
Thx David, Carlos Sanchez RiskMetrics -----Original Message----- From: owner-xsl-list@xxxxxxxxxxxxxxxx [mailto:owner-xsl-list@xxxxxxxxxxxxxxxx]On Behalf Of David Carlisle Sent: Tuesday, April 11, 2000 11:52 AM To: xsl-list@xxxxxxxxxxxxxxxx Subject: Re: XSL Copy > 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? tags don't have names (elements do) For the root node and almost all its descendents, you have not specified any template, so you get the default template which just recurses on the children and finally copies the character data to the result. <xsl:template match="//timeSeriesDetail/timeSeries[timeSeriesType=$tsType and timeSeriesKey=$tsName]"> using // is almost always inefficient, or just plain wrong, and never makes sense in a match pattern. What I think you want is <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="/"> <xsl:copy-of select="results/timeSeriesDetail[ timeSeries/timeSeriesType=$tsTyp and timeSeries/timeSeriesKey=$tsName]"/> </xsl:template> </xsl:stylesheet> but its hard to be sure... David XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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