xsl-list
[Top] [All Lists]

Re: super basic xsl question

2005-01-13 10:09:33
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(_at_)protosw(_dot_)com> 
01/13/2005 10:58 AM
Please respond to
xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com


To
xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
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.



--~------------------------------------------------------------------
XSL-List info and archive:  http://www.mulberrytech.com/xsl/xsl-list
To unsubscribe, go to: http://lists.mulberrytech.com/xsl-list/
or e-mail: <mailto:xsl-list-unsubscribe(_at_)lists(_dot_)mulberrytech(_dot_)com>
--~--




--~------------------------------------------------------------------
XSL-List info and archive:  http://www.mulberrytech.com/xsl/xsl-list
To unsubscribe, go to: http://lists.mulberrytech.com/xsl-list/
or e-mail: <mailto:xsl-list-unsubscribe(_at_)lists(_dot_)mulberrytech(_dot_)com>
--~--



<Prev in Thread] Current Thread [Next in Thread>