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Return position in for-each

Subject: Return position in for-each
From: "Karl J. Stubsjoen" <karl@xxxxxxxxxxxxxxxxxxxx>
Date: Thu, 22 Jul 2004 15:45:40 -0700
xsl position for each
Hello,

A disadvantage to an xsl:for-each is that the position() increases by one
for each step.  so, if an xsl:if test validates the current step and says
"Yes apply templates for this step", "don't apply templates for this step",
and you are interested in a positiion which reflects the total number of
"YES" steps... well you have a problem.  Take into consideration this
for-each rule:

<xsl:for-each select="Entry">
<xsl:variable name="element_name"><xsl:value-of
select="concat('competitor_name_',position())"/></xsl:variable>
  <xsl:if test="string($BROWSER_VARS//ELEMENT[@name=$element_name]/.)">
    <tr>

	<!-- [[[ HERE WE ADD A COLUMN THAT HAS AN INCREMENTING NUMBER ]]] -->
      <td><xsl:value-of select="concat(position(),'.')"/></td>

	<!-- [[[ HERE WE GET THE REST OF THE COLUMN DATA ]]] -->
      <xsl:apply-templates select="Element" mode="PRINTER_FRIENDLY">
         <xsl:with-param name="position"><xsl:value-of
select="position()"/></xsl:with-param>
      </xsl:apply-templates>
    </tr>
  </xsl:if>
</xsl:for-each>


Is there a work-around, or another way to accomplish an incrementing number
for the TD?

Karl

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