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Making balanced two-column tables from one-column data


Subject: Making balanced two-column tables from one-column data
From: Philip Borenstein <pb@xxxxxxxxxx>
Date: Wed, 19 Jan 2000 13:35:12 -0500

I have a solution to my problem, but it seems so un-XSLT-like, and ugly that
I'm sure there must be a better solution. i think that part of the problem
is that I just don't "get" XPaths. Pointers to good XPath learning resources
much appreciated.

I have the following XML:

<TASKS>
  <TASK>
    <COMPONENTS>
      <COMPONENT>A</COMPONENT>
      <COMPONENT>B</COMPONENT>
      <COMPONENT>C</COMPONENT>
      <COMPONENT>D</COMPONENT>
      <COMPONENT>E</COMPONENT>
      <COMPONENT>F</COMPONENT>
      <COMPONENT>G</COMPONENT>
      <COMPONENT>H</COMPONENT>
      <COMPONENT>I</COMPONENT>
      <COMPONENT>J</COMPONENT>
      <COMPONENT>K</COMPONENT>
    </COMPONENTS>
  </TASK>
</TASKS>

I want to produce a two-column table like this:

  A   G
  B   H
  C   I
  D  J
  E  K
  F

(The number of COMPONENTS may or may not be even.)

This is my XSL template for the COMPONENT element:

<xsl:template match="COMPONENTS">
  <P>There are <xsl:value-of select="count(COMPONENT)"/> components.</P>
  <TABLE>
     <xsl:variable name="h" select="round(count(COMPONENT) div 2)" />
     <xsl:for-each select="COMPONENT">
       <xsl:variable name="p" select="position()" /> 
       <xsl:if test="$h>=$p">
           <TR>
                  <TD><xsl:value-of select="../COMPONENT[position()=$p]"
/></TD>
                  <TD><xsl:value-of select="../COMPONENT[position()=$p+$h]"
/></TD>
            </TR>
         </xsl:if>
      </xsl:for-each>
  </TABLE>
</xsl:template>

This just strikes me as the wrong way to go about it.

--philip borenstein.


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