Subject: Re: [xsl] different first element in a list From: Lorenzo De Tomasi <detomasi.liste@xxxxxxxxx> Date: Mon, 24 Feb 2003 17:12:22 +0100 |
on 24-02-2003 10:07, Michael Kay at michael.h.kay@xxxxxxxxxxxx wrote: > In the inner for-each, instead of > > xsl:for-each select="key('k', @type)" > > try > > xsl:for-each select="key('k', @type)[position()!=1]" Now it works :) Thank you very much... but I still have a lot of problems in understanding how this works Can somebody help me to understand commenting the code? The Xslt I is at <http://biografica.tzone.it/cv/eucv_eng_prova.xsl>) Let's start from the basic concepts: If I have an xml list like this __________________________________________ Xml <xml> <element>1</element> <element>3</element> <element>2</element> <element>4</element> <element>5</element> </xml> __________________________________________ how can I obtain an ordered list like this __________________________________________ Xhtml <table> <tr> <td>list:</td> <td>1</td> </tr> <tr> <td/> <td>2</td> </tr> <tr> <td/> <td>3</td> </tr> <tr> <td/> <td>4</td> </tr> <tr> <td/> <td>5</td> </tr> </table> __________________________________________ Rendering list: 1 2 3 4 5 Thank you very much :) And excuse me if I don't understand, but I have not a programmer's brain (or I'm simply stupid :P) If you need the help of an information architect, please ask freely :) XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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