Re: [xsl] The longest node in a node set

[Jeni Tennison <jeni@xxxxxxxxxxxxxxxx>]

Hi Antonio,

> I am trying to find an XPath expression that gives me the longest
> element in a node set.

This is a variation on the "computed maximum" problem. You can't do it
within a single XPath, at least not in XPath 1.0, so you have to use a
different technique. Depending on how many elements you have, you
could use a sort-and-select method:

  <xsl:for-each select="//element">
    <xsl:sort select="string-length()"
              data-type="number"
              order="descending" />
    <xsl:if test="position() = 1">
      Longest: <xsl:value-of select="." />
    </xsl:if>
  </xsl:for-each>

or you could use a recursive template, by tailoring the one in
Dimitre's FXSL library or by writing your own, such as:

<xsl:template name="findLongest">
  <xsl:param name="nodes" select="//element" />
  <xsl:param name="max" select="0" />
  <xsl:param name="longest" select="/.." />
  <xsl:choose>
    <xsl:when test="$nodes">
      <xsl:variable name="length" select="string-length($nodes[1])" />
      <xsl:call-template name="findLongest">
        <xsl:with-param name="nodes"
                        select="$nodes[position() > 1]" />
        <xsl:with-param name="max">
          <xsl:choose>
            <xsl:when test="$max >= $length">
              <xsl:value-of select="$max" />
            </xsl:when>
            <xsl:otherwise>
              <xsl:value-of select="$length" />
            </xsl:otherwise>
          </xsl:choose>
        </xsl:with-param>
        <xsl:with-param name="longest"
                        select="$longest[$max >= $length] |
                                $nodes[1][$length >= $max]" />
      </xsl:call-template>
    </xsl:when>
    <xsl:otherwise>
      Longest: <xsl:value-of select="$longest" />
    </xsl:otherwise>
  </xsl:choose>
</xsl:template>

This template actually finds *all* the elements that are longest, in
case there are several with the same length, whereas using the
sort-and-select method you only get the first with that longest
length.

Cheers,

Jeni

---
Jeni Tennison
http://www.jenitennison.com/


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