Subject: Re: [xsl] (simple?) xpath question From: mozer <xmlizer@xxxxxxxxx> Date: Fri, 29 Aug 2008 22:08:23 +0200 |
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: [xsl] (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>
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