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RE: [xsl] Why no namespace node KindTest?
Subject: RE: [xsl] Why no namespace node KindTest? From: "Michael Kay" <mike@xxxxxxxxxxxx> Date: Sat, 26 Aug 2006 21:38:17 +0100 |
> Because namespace nodes are deprecated, and a compliant > processor need not support them. Not strictly true in the case of XSLT. Every processor is required to support the following: <xsl:variable name="x" as="node()"> <xsl:namespace name="xs">http://www.w3.org/2001/XMLSchema</xsl:namespace> </xsl:variable> <xsl:if test="$x instance of ?????">... What is true, however, is that if you don't support the namespace axis then you never need to expose a namespace node that's attached to an element, which means you don't get into the XPath 1.0 issues of namespace node identity. (The background on this is that people realized that namespace nodes were a lot more expensive than they needed to be because they had separate identity and separate parents, which makes it impossible for different elements to share namespace nodes. Since there are no known use cases for namespace nodes where the identity and parentage actually matter, use of the namespace axis was deprecated and replaced by the functions in-scope-prefixes and namespace-uri-for-prefix. XQuery is written as if namespace nodes don't exist, even though they are defined in the data model, because XQuery never exposes them. XSLT however relies on namespace nodes to support dynamic namespace construction using xsl:namespace.) Michael Kay http://www.saxonica.com/
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