Re: [xsl] Likely a very stupid question

[Joelle Tegwen <tegwe002@xxxxxxx>]

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