Re: [xsl] xml to xml transform

Subject: Re: [xsl] xml to xml transform
From: Jeni Tennison <mail@xxxxxxxxxxxxxxxx>
Date: Tue, 16 Jan 2001 10:32:54 +0000
Hi Shimon,

Here's the Muenchian way (see
very similar to the one Mike B. showed you, but uses keys to retrieve
the relevant 'row' elements and so will probably be more efficient if
you've got lots of rows.

<?xml version="1.0" encoding="utf-8"?>
<xsl:stylesheet xmlns:xsl="";

<xsl:output method="xml" indent="yes"/>

<xsl:key name="rows" match="row" use="@fld" />

<xsl:template match="rows">
   <xsl:for-each select="row[count(.|key('rows', @fld)[1]) = 1]">
      <xsl:element name="{@fld}">
         <xsl:for-each select="key('rows', @fld)">
            <xsl:element name="{@value}"/>


I hope that helps,


Jeni Tennison

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