xsl-list
[Top] [All Lists]

Re: [xsl] How to insert a set nodes under the root of an arbitrary XML using XSL?

2011-01-07 18:24:52
On Fri, Jan 07, 2011 at 05:00:56PM -0500, Wendell Piez scripsit:
Hi again,

On 1/7/2011 4:49 PM, was written:
The first match I gave is a match on the document node; / is the root
node, /* is the document node, at least per XPath 2 terminology.  XPath
1 is of course using different terminology.:)

Not quite. "/" is both "the root node" (when you know what tree
you're talking about" and "a document node" (when you don't).

/* (short for /child::*) is the "document element" of a well-formed
XML instance, but we also see trees (for example, it's not uncommon
in temporary trees) that have more than one element child of the
root; and this XPath will select all of them.

Argh.

Yes, of course.

Pardon me while I attempt to cudgel my brain into a state of greater
utility.

-- Graydon

--~------------------------------------------------------------------
XSL-List info and archive:  http://www.mulberrytech.com/xsl/xsl-list
To unsubscribe, go to: http://lists.mulberrytech.com/xsl-list/
or e-mail: <mailto:xsl-list-unsubscribe(_at_)lists(_dot_)mulberrytech(_dot_)com>
--~--