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@xxxxxxxxx> 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@xxxxxxxxx> wrote:
>
>> From: Xmlizer <xmlizer+xsllist@xxxxxxxxx>
>> Subject: Re:  (simple?) xpath question
>> To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
>> 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@xxxxxxxxx> 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>