Subject: Re: [xsl] Equivalence between XSL and XPath expression From: Dimtre Novatchev <dnovatchev@xxxxxxxxx> Date: Sat, 11 Dec 2004 12:12:36 +1100 |
On Sat, 11 Dec 2004 00:54:19 +0000, xptm <xptm@xxxxxxx> wrote: > So basically you're saying that the root element doesn't have the self:: > axis, besides the obvious ancestor, parent and preceding. Is that so? Any node, including "/" "has a self axis". However, the expression you suggested: count(./ancestor-or-self::*)+count(./preceding::*) evaluates to 0 in the case when the context node is the document node. The reason? The principal node kind for the self axis is the element-node kind. Therefore, self::* selects the context node only if the context node is an element. This is not the case with the root (document) node. Correct the above to: self::node() and it now selects the context node always, regardless of its node-kind. Cheers, Dimitre.
Current Thread |
---|
|
<- Previous | Index | Next -> |
---|---|---|
Re: [xsl] Equivalence between XSL a, xptm | Thread | Re: [xsl] Equivalence between XSL a, xptm |
Re: [xsl] Equivalence between XSL a, xptm | Date | Re: [xsl] Equivalence between XSL a, xptm |
Month |