Re: Grouping and conditional info questions

Hi,  Ying Qin

The following xslt may be one of your solutions.

<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
  <xsl:output method="xml"  indent="yes"/>
<xsl:key name="item" match="item" use="id" />

<xsl:template match="/">
<groups>
  <xsl:for-each
      select="docs/item[count(.|key('item', id)[1]) = 1]">
<group>
    <title><xsl:text>Group</xsl:text>
      <xsl:number value="position()" format=" A"/>
      </title>
    <xsl:for-each select="key('item', id)">
     <item>
      <xsl:copy-of select="*"/>
     </item>
    </xsl:for-each>
    <xsl:text>&#xA;</xsl:text>
</group>
  </xsl:for-each>
</groups>
</xsl:template>
</xsl:stylesheet>

Basically,  your problem is a typical group method, which
Jeni Tennison is one of the best to illustrate the solution for this kind
of method.

One of her most most recent post is on
http://sources.redhat.com/ml/xsl-list/2001-08/msg01289.html(title:grouping
headers)
I highly recommend you to read this one  and her web site. It will help
yourself a
lot to be familiar with the subject about how to solve the group problem
using key and
Muenchian method.

Besides, copy-of is to copy  child node of item,  and xsl:number to get your
desired title name.

Hope it will help.

cheers.


Sun-fu Yang,

sfyang@xxxxxxxxxxxxx


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