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<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="2.0">
<xsl:param name="path" select="'/part[1]'"/> <!-- default value -->
<xsl:template match="$path">
<xsl:apply-templates mode="copy"/>
</xsl:template>
<xsl:template match="@*|node()" mode="copy">
<xsl:copy>
<xsl:copy-of select="@*"/>
<xsl:apply-templates mode="copy"/>
</xsl:copy>
</xsl:template>
</xsl:stylesheet>
What is the best way of going about this? Thanks.
[xsl] Copy path specified by parameter
Subject: [xsl] Copy path specified by parameter From: "Steve Lloyd" <steve.lloyd@xxxxxxx> Date: Thu, 14 Jul 2005 14:04:23 +0100 |
I'd like to return a sub-path which is selected by a parameter to the XSLT. Something like:-
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="2.0">
<xsl:param name="path" select="'/part[1]'"/> <!-- default value -->
<xsl:template match="$path">
<xsl:apply-templates mode="copy"/>
</xsl:template>
<xsl:template match="@*|node()" mode="copy">
<xsl:copy>
<xsl:copy-of select="@*"/>
<xsl:apply-templates mode="copy"/>
</xsl:copy>
</xsl:template>
</xsl:stylesheet>
What is the best way of going about this? Thanks.
-- Steve Lloyd XML Developer, Publications group, International Baccalaureate Organization Email: steve.lloyd[at]ibo.org Tel: +44 (29) 2054 7869 WWW: http://www.ibo.org
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