Subject: [xsl] Re: Re: "Instantiating a template" ? From: "Dimitre Novatchev" <dnovatchev@xxxxxxxxx> Date: Tue, 2 Dec 2003 23:13:00 +0100 |
"David Tolpin" <dvd@xxxxxxxxxxxxxx> wrote in message news:200312022033.hB2KXmfr061117@xxxxxxxxxxxxxxxxxxxxxx > > I think you're not making difference between a match pattern and an XPath > > expression. Match paterns are only a restricted subset of the set of > > syntactically correct XPath expressions. > > > > As Wendell pointed out, this distinction is at the heart of XSLT. > > Dimitre, > > I was commenting on the exact cause due to which node() cannot match the root > node. I was trying to show that the cause is not that node() is required to be > a child of something, as another poster had suggested, but is that node() is a > short way to write child::node(). Yes, what you said was: > Because not specifying an axis is the same as specifying child:: axis. > There is no context in wich the root node resides on the child axis. Yes. > > self::node() in the context of the root element will match the root node. -----------------------------------------------------------^^^^ In XSLT the verb "to match" is used with match patterns. Because only the "child" and "attribute" axis are allowed in a match pattern (except within predicates), the expression: self:node cannot *match* anything as it simply is not allowed as a match pattern. Another problem with this statement is that it uses the term "root element", but the root node in fact is not an element node. The expression self::node() in the context of (any) element will never select the root node of the document. What *is* true is that the above expression in the context of the root node (not the root element) will *select* (not match) the root node. ===== Cheers, Dimitre Novatchev. http://fxsl.sourceforge.net/ -- the home of FXSL XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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