Re: Reformatting tables

<xsl:template match="tr">
    <xsl:for-each select="td[1]/p">
        <xsl:variable name="pos" select="position()"/>
        <tr>
            <td><xsl:copy-of select="."/></td>
            <xsl:for-each select="../following-sibling::td">
                <td><xsl:copy-of select="p[$pos]"/></td>
            </xsl:for-each>
        </tr>
    </xsl:for-each>
</xsl:template>

Hi all,

I have a large number of html tables in the following format:

<table>
  <tr>
    <td>
      <p>Animal</p>
    </td>
    <td>
      <p>Colour</p>
    </td>
  </tr>
  <tr>
    <td>
      <p>Elephant</p>
      <p>Pig</p>
      <p>Sheep</p>
      <p>Canary</p>
    </td>
    <td>
      <p>Grey</p>
      <p>Pink</p>
      <p>White</p>
      <p>Yellow</p>
    </td>
  </tr>
</table>

Where someone has used FrontPage or something and has ended up creating one
row of multiple paragraphs instead of a row for each item.

How can I write some xsl to reformat this into proper rows? Eg:

<tr>
  <td>
   <p>Elephant</p>
  </td>
  <td>
   <p>Grey</p>
  </td>
</tr>

So far I have made:

<xsl:template match="tr">
 <xsl:for-each select="td[1]/p">
  <tr>
   <td><p><xsl:value-of select="."/></p></td>
   <xsl:variable name="n" select="position()"/>
   <td><p><xsl:value-of select="//td[2]/p[position()=$n]"/></p></td>
  </tr>
 </xsl:for-each>
</xsl:template>

But this always leaves them out of sync.

What am I doing wrong?


Thanks in advance,


Neil.

PS. Excuse the animal colours example, the real data is confidential
internal stuff :-)

System Development
VIRBUS AG
Fon  +49(0)341-979-7411
Fax  +49(0)341-979-7409
joerg.heinicke@xxxxxxxxx
www.virbus.de

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