Subject: Re: [xsl] What is the simplest method for using xsl:sort without losing attribute names and values? From: Martin Honnen <Martin.Honnen@xxxxxx> Date: Mon, 21 Jul 2008 13:55:41 +0200 |
In the example below, only one <Item> is shown. I tried the template you suggested.
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0">
<xsl:strip-space elements="*"/> <xsl:output method="xml" indent="yes"/>
<xsl:template match="@* | node()"> <xsl:copy> <xsl:apply-templates select="@* | node()"/> </xsl:copy> </xsl:template>
<xsl:template match="List"> <xsl:copy> <xsl:apply-templates select="@*"/> <xsl:apply-templates> <xsl:sort select="Article/Year" /> <xsl:sort select="Article/IssueNumber"/> <xsl:sort select="Article/Page" /> </xsl:apply-templates> </xsl:copy> </xsl:template>
Martin Honnen http://JavaScript.FAQTs.com/
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