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Re: [xsl] super basic xsl question


Subject: Re: [xsl] super basic xsl question
From: Jeb Boniakowski <jeb@xxxxxxxxxxx>
Date: Thu, 13 Jan 2005 12:06:53 -0500

Thanks a lot, that's what I was looking for.

jeb..

On Jan 13, 2005, at 12:09 PM, JBryant@xxxxxxxxx wrote:

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.


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