Subject: Re: [xsl] Position() of parent node From: Joe English <jenglish@xxxxxxxxxxxxx> Date: Tue, 06 Feb 2001 13:41:59 -0800 |
Simon Cansick wrote: > Can anyone provide me with the syntax for getting the position() value of > the current nodes' parent node (and the parent parent etc. position() > value). I seem only able to return the current position(). The position() is not an intrinsic property of a node -- it only makes sense to ask for the position() of a node _in the context of some node list_ [*]. Most likely you want the position of the parent node with respect to its siblings, in document order; you can compute this with: 1 + count(parent::*/preceding-sibling::*) --Joe English jenglish@xxxxxxxxxxxxx [*] I'm using the term "node list" to mean "a node set with an associated order", e.g., proximity position order, document order, an <xsl:sort...>-defined order, etc. XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
Current Thread |
---|
|
<- Previous | Index | Next -> |
---|---|---|
Re: [xsl] Position() of parent node, David Carlisle | Thread | [xsl] Same problem: Unknown op code, Matthias O. Will |
Re: [xsl] XInclude in Cocoon, Bill Humphries | Date | Re: [xsl] XInclude in Cocoon, Paul Grosso |
Month |