Re: [xsl] Likely a very stupid question

Subject: Re: [xsl] Likely a very stupid question
From: Joelle Tegwen <tegwe002@xxxxxxx>
Date: Fri, 20 Oct 2006 12:52:42 -0500
At this point I'm just trying to transform my xml page. Eventually I'll probably want to do a nested list or something like that,

Even when I add
   <table border="1">
     <tr bgcolor="#9acd32">
       <th>ID</th>

       <th>Title</th>
     </tr>
     <xsl:for-each select="sitemap/section">
     <tr>
       <td><xsl:value-of select="id" /></td>
       <td><xsl:value-of select="title" /></td>
     </tr>
     </xsl:for-each>
   </table>

I get nothing. I would at least expect to get one section or two in my table. But maybe I don't understand enough about what's going on.
Thanks
J


Mukul Gandhi wrote:
You seem to have missed a xsl:for-each or xsl:apply-templates instruction.

In your Schema, there is recursive nesting of <section> tag. So you
need to show us the desired output as well (for the example you have
given).

On 10/20/06, Joelle Tegwen <tegwe002@xxxxxxx> wrote:
I'm just trying to figure out XML and XSL.
I've been using this tutorial
http://www.w3schools.com/xsl/xsl_value_of.asp to guide me through as I
work on my own code.

I've got this http://rafb.net/paste/results/ObLbVH51.html schema
for this http://rafb.net/paste/results/dyNo4Q36.html xml file
and I'm trying to apply this http://rafb.net/paste/results/SOetCN41.html
xsl stylesheet

I get no values in my table, just the headers. I've even tried copying
and pasting their code and just entering the corresponding code for my
own document and I still get nothing.

What am I doing wrong?

Thanks
Joelle

Current Thread