Subject: [xsl] Can you pass a parameter from a stylesheet to the resulting HTML file? From: Grant-Kathryn@xxxxxxxxxxxxxxxxx Date: Thu, 30 May 2002 17:56:19 -0700 |
Hi all, I sure appreciate everyone's help! The manual I'm working on is going well because of it. I have a more complicated question. Part of my XHTML file contains a table, simplified below: <table> <tr> <td> <ol> <li>item 1</li> <li>item 2</li> <li>item 3</li> </ol> </td> </tr> <tr> <td> <ol start="4"> <li>item 4</li> <li>item 5</li> <li>item 6</li> </ol> </td> </tr> </table> As you can see, I have a continuous list split across two cells. Even if I don't close the </ol> in the first cell, the numbering starts again at 1 in the next cell unless I use <ol start="4">. My question is this. I have two different stylesheets transforming the same XHTML file. One stylesheet filters the XHTML file so the number of list items is less; therefore, the <ol> start attribute needs to be different in each transformed file. I understand that I can declare a parameter in my XSL file thus: <xsl:param name="startnum" select="3"/> But is there a way to pass that parameter to the <ol> tag's start attribute in the transformed file? I've checked my XSL books and searched on the web, but I can't find a way to do this. Below is the XSL file I'm using--it's pretty simple. ----------------------xsl file--------------------------- <?xml version="1.0" encoding="UTF-8"?> <xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:xhtml="http://www.w3.org/1999/xhtml" version="1.0"> <xsl:output method="xml" encoding="UTF-8"/> <xsl:template match="node() | @*"> <xsl:copy> <xsl:apply-templates select="node() | @*"/> </xsl:copy> </xsl:template> </xsl:stylesheet> --------------------end xsl file--------------------------- TIA, Kathryn XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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