Subject: [xsl] Alternative to variable in template match From: Rick Anderson <rianders@xxxxxxxxxxxxxxx> Date: Thu, 17 Jan 2002 15:54:28 -0500 (EST) |
I wrote a small xsl to copy the nodes below a specified XPATH expression into new files. I wanted to eventually pass the XPATH in via a command line. Since a variable can't be used in 'template match="$path"' there must be another efficient way to do this. Currently, my work around doesn't compare the XPATH expression to the node successfully. Any hints? xsl: ---- <xsl:stylesheet version="1.1" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"> <xsl:output method="xml"/> <xsl:variable name="dir" select="'out'" /> <xsl:variable name="prefix" select="'name'" /> <xsl:variable name="suffix" select="'xml'" /> <xsl:variable name="xpath" select="a"/> <xsl:template match="*"> <xsl:if test="$xpath" > <xsl:variable name="filename" select="concat($dir,'/',$prefix, position(),'.',$suffix)"/> <xsl:document href="{$filename}"> <doc> <xsl:copy-of select="node()"/> </doc> </xsl:document> </xsl:if> </xsl:template> </xsl:stylesheet> ---- xml: ---- <doc> <a> <b> <c>This is 1</c> </b> </a> <a> <b> <c>This is 2</c> </b> </a> <a> <b> <c>This is 3</c> </b> </a> </doc> ---- Thanks, --Rick Anderson XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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