Once I start thing of the stylesheet as a xml document I get it,
document('') is the root and the rest follow..
thanks,
-RVG
----- Original Message -----
From: "Dimitre Novatchev" <dnovatchev(_at_)yahoo(_dot_)com>
To: <xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com>
Sent: Saturday, November 22, 2003 6:53 PM
Subject: [xsl] Re: refering to internal xml data
"Robert Van Gemert" <rcvangemert(_at_)optushome(_dot_)com(_dot_)au> wrote in
message
news:004101c3b08a$ece10090$51011f0a(_at_)CO3050560A(_dot_)(_dot_)(_dot_)
Thanks David,
I presume document('') refers to a null document i'll look more into
this
syntax..
document('')
selects the root of the current xsl:stylesheet document. E.g:
document('')/*
selects the xsl:stylesheet element.
document('')/*/xsl:variable
selects all global xsl:variable (-s),
document('')/*/xsl:template
selects all xsl:template (-s),
etc.
=====
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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