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(_dot_)Elovirta(_at_)nokia(_dot_)com
[mailto:Jarno(_dot_)Elovirta(_at_)nokia(_dot_)com]
Sent: Monday, April 28, 2003 5:31 PM
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
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
XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list