RE: [xsl] Vertical display

Subject: RE: [xsl] Vertical display
From: Sundar Shanmugasundaram <SSHANMUGASUNDARAM@xxxxxxxxxxxxx>
Date: Mon, 28 Apr 2003 18:08:00 +0530
Jarno,

Sorry, It didn't properly, when i give
this xml as input.

<o>
	<com>
		<hereyougo>
			<first>1</first>
			<second>2</second>
			<third>3</third>

		</hereyougo>
		<imaycome>
			<four>4</four>
			<five>5</five>
			<six>6</six>

		</imaycome>

	</com>
</o>

Note here it outputs as follows:

1 4 
2 5 

I think this is the problem with assigning position to outer node.

How to solve this?

sundar

-----Original Message-----
From: Jarno.Elovirta@xxxxxxxxx [mailto:Jarno.Elovirta@xxxxxxxxx]
Sent: Monday, April 28, 2003 5:31 PM
To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
Subject: RE: [xsl] Vertical display


Hi,

> This is the xml file. I want to display in HTML table as follows :
> 
> 1      4        41       42
> 2      5        51       52
> 3      6        61       62
> 
> How will i do that?
> 
> Please give XSL template.

<xsl:template match="/">
  <html>
    <head>
      <title/>
    </head>
    <body>
      <table>
        <tbody>
          <xsl:for-each select="o/com/*">
            <xsl:variable name="x" select="position()" />
            <tr>
              <xsl:for-each select="../*/*[position() = $x]">
                <td>
                  <xsl:value-of select="." />
                </td>
              </xsl:for-each>
            </tr>
          </xsl:for-each>
        </tbody>
      </table>
    </body>
  </html>
</xsl:template>

I suppose there's an entry for reversing a table in the XSLT FAQ, but I
didn't find it. If the number of cells rows varies, first calculate the max
number of rows and then process accordingly--I think I posted a solution for
it earlier this month.

Cheers,

Jarno - God Module: Telekinetic

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