that's not possible to do with XPath only
But with XSLT you can
Xmlizer
On Fri, Aug 29, 2008 at 9:06 PM, mark bordelon
<markcbordelon(_at_)yahoo(_dot_)com> wrote:
All *help*!
What is the best way to query xml with xpath to get a disjoint nodelist?
Specifically i want to include just the root node alongwith a descendent node.
XML:
<a>
<b>
<c>
</c>
</b>
</a>
XPATH:
//c
DESIRED RESULT NODELIST:
i.e. not this:
<c>
</c>
but rather this:
<a>
<c>
</c>
</a>
--~------------------------------------------------------------------
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>
--~--