Re: [xsl] super basic xsl question

[JBryant@xxxxxxxxx]

Don't have time for a long reply, but try xsl:copy-of

Jay Bryant
Bryant Communication Services
(on contract at Synergistic Solution Technologies)




Jeb Boniakowski <jeb@xxxxxxxxxxx> 
01/13/2005 10:58 AM
Please respond to
xsl-list@xxxxxxxxxxxxxxxxxxxxxx


To
xsl-list@xxxxxxxxxxxxxxxxxxxxxx
cc

Subject
[xsl] super basic xsl question






Thanks to everyone on the list that helped me with my last problem.  I 
still have some fundamental misunderstanding of how xsl is supposed to 
work, because I can't figure out how to do this:

<xmlroot>
  <child>some text</child>
  <child><a href="">a link</child>
</xmlroot>

and convert it to say...

<ul>
  <li>some text</li>
  <li><a href="">a link</child>
</ul>

I've tried various combos like:
<xsl:template match="child">
                 <xsl:value-of select="."/>
</xsl:template>

but that means I lose the <a> tags.

I thought something with <xsl:copy> would work, but that gives me the 
context node, and furthermore, it doesn't give me the children, so it 
does the opposite.  Then I thought fiddling with the 'select' expr in 
the value-of tag would do it, but I can't figure out if there's a 
magical combination of XPath slang that means, "whatever the hell is 
below here, be it tags or text, i want them".

jeb.