RE: How to unflatten the source XML ?

Hi,

> Would you please tell me how to group fixed number (say 3) of 
> consecutive node together?
  <xsl:template match="*[item]">
    <xsl:for-each select="item[position() mod 3 = 1]">
      <group id="{position()}">
        <xsl:copy-of select=". | following-sibling::item[position() &lt; 3]"/>
      </group>
    </xsl:for-each>
  </xsl:template>

Cheers,

Jarno - Project Pitchfork: Carrion

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