Re: grouping list items by attribute

Please try this stylesheet

<?xml version="1.0"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0">

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

<xsl:template match="/tx.list">
   <ol>
     <xsl:for-each select="tx.li">
       <xsl:choose>
         <xsl:when test="@level = '0'">
           <li><xsl:value-of select="tx.p" /></li>
         </xsl:when>
         <xsl:otherwise>
           <xsl:if test="(position() = 1) or
preceding-sibling::tx.li[1]/@level = '0'">
             <ul>
               <li><xsl:value-of select="tx.p" /></li>
               <xsl:call-template name="printli">
                 <xsl:with-param name="list"
select="following-sibling::tx.li" />
               </xsl:call-template>
             </ul>
           </xsl:if>
         </xsl:otherwise>
       </xsl:choose>
     </xsl:for-each>
   </ol>
</xsl:template>

<xsl:template name="printli">
   <xsl:param name="list" />

   <xsl:if test="$list/@level = '1'">
     <li><xsl:value-of select="tx.p" /></li>
     <xsl:call-template name="printli">
       <xsl:with-param name="list" select="$list[position() &gt; 1]" />
     </xsl:call-template>
   </xsl:if>
</xsl:template>

</xsl:stylesheet>

Regards,
Mukul

On 10/4/05, Lynn Alford <lynn.alford@xxxxxxxxxx> wrote:
> Given this as the input
>
> <tx.list style="ID1c6">
>           <tx.li level="0" number="1.   ">
>             <tx.p style="ID1a2">List1</tx.p>
>           </tx.li>
>           <tx.li level="0" number="2.   ">
>             <tx.p style="ID1a2">List2</tx.p>
>           </tx.li>
>           <tx.li level="1" number="o?=    ">
>             <tx.p style="ID1a2">Sublist 1</tx.p>
>           </tx.li>
>           <tx.li level="1" number="o?=    ">
>             <tx.p style="ID1a2">Sublist 2</tx.p>
>           </tx.li>
>           <tx.li level="1" number="o?=    ">
>             <tx.p style="ID1a2">Sublist 3</tx.p>
>           </tx.li>
>           <tx.li level="0" number="3.   ">
>             <tx.p style="ID1a2">List3</tx.p>
>           </tx.li>
>         </tx.list>
>
> The preferred outcome would be
>
> <ol>
>           <li>List1</li>
>           <li>List2</li>
>           <ul>
>                   <li>Sublist 1</li>
>             <li>Sublist 2</li>
>             <li>Sublist 3</li>
>           </ul>
>           <li>List3</li>
>         </ol>
>
> My basic problem is I can't quite figure out how you can use an increase in
> level to trigger the start of a new list.  Especially when the lists may be
> nested much deeper and any level may be either ordered or unordered
> lists.  Siblings at the same level I can do easily enough.
>
> Lynn
>
> Lynn Alford                             Tel     (07) 47 81 6256
> ITR                                             Email:  imla@xxxxxxxxxx
> JCU QLD 4811 Australia                  ICQ: 64096907
> MSN: nicarra60@xxxxxxxxxxx              Y!: nicarra60
>
> 'I think an "uncatched exception" deserves three bug reports: one for the
> exception, one for not catching it, and one for abuse of the English
> language.'  Michael Kay