Re: [xsl] Loading XML with XSL for sorting in ASP

Subject: Re: [xsl] Loading XML with XSL for sorting in ASP
From: Bjorn Van Blanckenberg <bjornvb@xxxxxx>
Date: Tue, 24 Jan 2006 16:07:58 +0100
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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