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 XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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