Subject: [xsl] is this really tough? From: Sundar Shanmugasundaram <SSHANMUGASUNDARAM@xxxxxxxxxxxxx> Date: Wed, 30 Apr 2003 15:03:48 +0530 |
I would like to sort the following XML file based on the displayposition. I am printing the value and displayname. Here is the XML File I have used: <?xml version="1.0"?> <o> <com> <hereyougo> <first displayposition="4" displayname="raman4">4</first> <second displayposition="1" displayname="raman1">1</second> <third displayposition="3" displayname="raman3">3</third> <fourth displayposition="2" displayname="raman2">2</fourth> </hereyougo> <imaycome> <four displayposition="4" displayname="raman4">22</four> <five displayposition="1" displayname="raman1">33</five> <six displayposition="3" displayname="raman3">44</six> <firvi displayposition="2" displayname="raman2">11</firvi> </imaycome> </com> </o> This is the output I expect: raman1 1 33 raman2 2 11 raman3 3 44 raman4 4 22 The o/p I am getting: raman1 4 22 raman2 1 33 raman3 3 44 raman4 2 11 The XSL File I used to process This : <?xml version="1.0"?> <xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"> <xsl:output method="html" indent="yes" /> <xsl:template match="/"> <table> <xsl:variable name="y" select="o/com/*" /> <xsl:for-each select="$y[1]/*"> <xsl:sort select="@displayposition"/> <xsl:variable name="x" select="position()" /> <tr> <th> <xsl:value-of select="@displayname"/> </th> <xsl:for-each select="$y/*[position() = $x]"> <xsl:sort select="@displayposition"/> <td> <xsl:value-of select="." /> </td> </xsl:for-each> </tr> </xsl:for-each> </table> </xsl:template> </xsl:stylesheet> Possible Error: <xsl:sort select="@displayposition"/> in the second <xsl:for-each > is not working. Pls help me out. thanks and regards, sundar XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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