Subject: Re: [xsl] better way to get the path to a node? From: Graydon <graydon@xxxxxxxxx> Date: Sat, 1 Dec 2012 08:48:39 -0500 |
On Sat, Dec 01, 2012 at 11:28:28AM +0100, Martin Honnen scripsit: > Graydon wrote: > >If I want to return the XPath path to a specific node when that node is > >the context node, is there a better way than: > > > ><xsl:sequence > >select="string-join(ancestor-or-self::*/concat('/',name(),'[',for $x in . return count(preceding-sibling::*[name() eq $x/name()])+1,']'),'')"/> > > > >"Better" here means "more efficient"; I'll be using various Saxon 9.* > >for this, either in oXygen or from java. > > > With Saxon there is > http://www.saxonica.com/documentation/extensions/functions/path.xml. > I haven't checked exactly however for which versions of Saxon 9.* it > is available. Thank you! I haven't, either, but can burn that bridge on Monday. (and check that saxon:path() really is faster for the data-set in question than the above. I'll be really surprised if it's not.) -- Graydon
Current Thread |
---|
|
<- Previous | Index | Next -> |
---|---|---|
Re: [xsl] better way to get the pat, Martin Honnen | Thread | Re: [xsl] better way to get the pat, Dimitre Novatchev |
Re: [xsl] better way to get the pat, Graydon | Date | Re: [xsl] better way to get the pat, Dimitre Novatchev |
Month |