Re: [newbie]use of xsl:if {RE: XSL to handle display mutiple pages}

[Mike Brown <mike@xxxxxxxx>]

My solution was the same as Jeni's. We had different ways of selecting
which item starts each page, and getting the next ($maxItemsPage -1)
items, but the overall method and outcome is the same. I also had a couple
of minor errors in my untested code sample. The corrected version is
below.

 <!-- start a new table for item 1, 51, 101, 151, etc. -->
 <xsl:for-each select="$items[(position()+1) mod $maxItemsPage + 1 = 1]">
   <xsl:variable name="currentPos" select="(position() - 1) * $maxItemsPage + 1"/>
   <table width="100%" border="1" cellspacing="0">
     <tbody>
       <!-- start a new row for current item and the next 49 -->
       <xsl:for-each select="$items[position() &gt;= $currentPos and position() &lt; $currentPos + $maxItemsPage]">
         <tr>
           <td>
             <xsl:value-of select="."/> <!-- or whatever -->
           </td>
         </tr>
       </xsl:for-each>
     </tbody>
   </table>
   <!-- output separator if we're not on the last item in the set -->
   <xsl:if test="$currentPos + $maxItemsPage &gt; count($books)">
     <div style='page-break-before: always'/>
   </xsl:if>
 </xsl:for-each>

   - Mike
____________________________________________________________________
Mike J. Brown, software engineer at         My XML/XSL resources:
webb.net in Denver, Colorado, USA           http://www.skew.org/xml/


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