Subject: Re: Loading XML with XSL for sorting in ASP
From: Bjorn Van Blanckenberg <bjornvb@xxxxxx>
Date: Tue, 24 Jan 2006 16:07:58 +0100
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This does it
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/
Transform">
<xsl:output method="xml" version="1.0" encoding="UTF-8" indent="yes"/>
<xsl:template match="/DVD">
<xsl:apply-templates>
<xsl:sort select="title" />
</xsl:apply-templates>
</xsl:template>
<xsl:template match="node()|@*">
<xsl:copy>
<xsl:apply-templates select="node()|@*">
<xsl:sort select="title" />
</xsl:apply-templates>
</xsl:copy>
</xsl:template>
</xsl:stylesheet>
How can I insert a second sort in /DVD/others by name with the same
xsl or do I have to use a second xsl
Thanks
On 24-jan-06, at 15:10, David Carlisle wrote:
The output xml I hope to get
<?xml version="1.0" encoding="ISO-8859-1"?>
<catalog>
<settings>
Your stylesheet is not generating any catalog or settings elements.
<xsl:template match="/catalog">
<xsl:apply-templates>
<xsl:sort select="title" />
</xsl:apply-templates>
</xsl:template>
This template doesn't generate an element it just applies templates to
the children of catalog.
You could add <xsl:copy or <catalog> but then it would be
equivalent to
your template matching node() so simplest is just to delete this
template.
David
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