Subject: Re: [xsl] (simple?) xpath question From: mark bordelon <markcbordelon@xxxxxxxxx> Date: Fri, 29 Aug 2008 13:01:18 -0700 (PDT) |
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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