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RE: no wonder i'm confused about the node() functio n
- From: Jeff Beal <jeff dot beal at ansys dot com>
- To: "'Robert P. J. Day'" <rpjday at mindspring dot com>,docbook apps list <docbook-apps at lists dot oasis-open dot org>
- Date: Mon, 03 Mar 2003 09:07:17 -0500
- Subject: RE: DOCBOOK-APPS: no wonder i'm confused about the node() functio n
Not having Tidwell's book, I can't comment on that. Michael Kay's
statement, though, is completely in line with the specification. The key
explanation is a few lines before what you quoted: "As an expression
<<node()>> is short for <<./child::node()>>, and the only nodes that this
can select are nodes that are children of something." If you were to write
a template expressly using the namespace axis (e.g. <xsl:template
match="namespace::node()"/>) You could use the node() "NodeTest" to match a
namespace node.
HTH,
Jeff Beal
-----Original Message-----
From: Robert P. J. Day [mailto:rpjday at mindspring dot com]
Sent: Sunday, March 02, 2003 3:02 PM
To: docbook apps list
Subject: DOCBOOK-APPS: no wonder i'm confused about the node() function
"XSLT", Doug Tidwell, p. 51
"* The node() node test, which selects all nodes in the current
context, regardless of type. This includes elements, text, comments
processing instuctions, attributes, and namespace nodes."
"XSLT Programmer's Reference", 2nd ed., Michael Kay, p. 432
"Since root nodes, attribute nodes and namespace nodes are never
children of another node ... they will never be matched by the
pattern node()."
"http://www.w3.org/TR/xslt"
"* node() matches any node other than an attribute node and
the root node"
at this point, i'm scared to read anything else for the fear
of getting a *fourth* opinion.
rday