Mark,
I think I was misunderstanding your question, because I undestood that
you wanted to construct a tree as a result
The fact is that XmlDocument.SelectNodes() only takes XPath
So you should follow the path drawn by Colin
Xmlizer
On Fri, Aug 29, 2008 at 10:01 PM, mark bordelon
<markcbordelon(_at_)yahoo(_dot_)com> wrote:
Xmlizer, thanks for the quick response!
Following your post, is there a simple enough xsl solution that could be
passed to
XmlDocument.SelectNodes(xslstr) to acheive the nodelist I want?
--- On Fri, 8/29/08, Xmlizer <xmlizer+xsllist(_at_)gmail(_dot_)com> wrote:
From: Xmlizer <xmlizer+xsllist(_at_)gmail(_dot_)com>
Subject: Re: [xsl] (simple?) xpath question
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Date: Friday, August 29, 2008, 12:12 PM
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>
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