Re:Reformatting a flat XML doc into an XML hierarchy b

Hi, Vidhya,
Following may help for your problem.
Please refer to recent Jeni's  reply for multi-level group
http://sources.redhat.com/ml/xsl-list/2001-04/msg00120.html

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output indent="yes"/>
<xsl:key name="foo"   match="row"  use="year"/>
<xsl:key name="bar"     match="row" use="concat(year,' ',acct)"/>

 <xsl:template match="/">
   <data>
     <xsl:for-each select="data/row[generate-id(.)
=generate-id(key('foo',year)[1])]">
 <xsl:variable name="year" select="year"/>
 <ActivityDate year="{$year}">
  <xsl:for-each select="key('foo',year)[generate-id(.)
=generate-id(key('bar',concat(year,' ',acct))[1])]">
 <xsl:variable name="acct" select="acct"/>
   <Acct id="{$acct}">
    <xsl:for-each select="key('bar',concat(year,' ',acct))">
      <Name>
       <xsl:value-of select="acctname"/>
      </Name>
     </xsl:for-each>
     </Acct>
   </xsl:for-each>
  </ActivityDate>
   </xsl:for-each>
   </data>
 </xsl:template>
</xsl:stylesheet>

hope it will help

sun-fu yang

sfynag@xxxxxxxxxxxxx



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