Re: Getting a longest node

Subject: Re: Getting a longest node
From: Sebastian Rahtz <sebastian.rahtz@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx>
Date: Mon, 17 Jan 2000 16:27:31 +0000 (GMT)
Wendell Piez writes:

 > Does anyone know a way I could define a variable that would contain the
 > number of characters in the longest node in a node-set? Let the node set in
 > question be //DIV[@type='Chapter']: if I have three, with string lengths
 > 88888, 99999, and 111110, I want my variable to be 111110.

I did something like this;  put all the values into a RTF,
convert it to a node-set, sort it numerically, and find the
last one. I append my (uncommented) code which attempts to work out
column widths for a table by adding together all the cell lengths in
each column and working out proportions


<xsl:template match="table">
<xsl:variable name="What">
   <xsl:value-of select="@id"/>
<record id="{$What}">
<xsl:variable name="cells">
 <xsl:for-each select=".//cell">
   <xsl:variable name="x"><xsl:number/></xsl:variable>
   <cell col="{$x}"><xsl:value-of select="string-length(.)"/></cell>
<xsl:variable name="total">
  <xsl:value-of select="sum(xt:node-set($cells)/cell)"/>
<xsl:for-each select="xt:node-set($cells)/cell">
  <xsl:sort select="@col" data-type="number"/>
  <xsl:variable name="c" select="@col"/>
  <xsl:if test="not(preceding-sibling::cell[$c=@col])">
   <xsl:variable name="len">
    <xsl:value-of select="sum(following-sibling::cell[$c=@col]) + current()"/>
   <fo:table-column column-number="{@col}" column-align="L" column-width="{$len div $total}" />

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