RE: [xsl] Select entire XML doc

[Wendell Piez <wapiez@xxxxxxxxxxxxxxxx>]

Hi Brian,

At 12:24 PM 3/4/2003, you wrote:
> Normally all I would want when doing
> a copy are elements.  I didn't realize that you *can* copy all nodes in a
> source document, probably because there's little point in doing so since you
> would just create an exact copy of the source.

Probably the commonest example in my experience is the technique of 
adapting an identity stylesheet to tweak a document only in places.

>   Perhaps if you were copying
> multiple sources to the result tree and wanted everything, including the
> <?xml?> declaration--but then the result wouldn't be well-formed XML,
> correct?

Yes -- with one tiny (but potentially important) caveat. The XML 
declaration <?xml version="1.0"?> is *not* a processing instruction, and is 
*not* present in the XPath data model. So in a document that starts:

<?xml version="1.0" encoding="funkyencoding"?>
<?xml-stylesheet href="myfunkyxform.xsl" type="text/xsl"?>

the xml-stylesheet PI is available, but the XML declaration (which looks 
just like a PI) is not -- nor is the information stored in it (such as that 
this data purportedly started life in some funkyencoding).

Why does the XML declaration look like a PI? (I hear you cry) if it's not?

One answer: for backwards compatibility with older (SGML) parsers that 
would not complain on a PI, but would gag if handed a declaration they 
didn't recognize. (The usual XML declaration syntax being, in conformance 
to SGML practice, <!...> .) Once upon a time this was a very handy 
workaround; it's still occasionally useful sometimes (at the cost of 
terrible confusion to parser writers or programmers who want "the XML 
processing instruction").

Cheers,
Wendell


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Wendell Piez                            mailto:wapiez@xxxxxxxxxxxxxxxx
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