Hi Pierre,
The answer is no in the general case. Consider for instance
<test>
<a>a</a>
<b>b</b>
</test>
<?xml version="1.0" encoding="UTF-8" ?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform";
version="1.0">
<xsl:template match="/">
result from which you cannot get the initial XML document
</xsl:template>
</xsl:stylesheet>
But you can get some information about the document and its content if you
look inside the stylesheet and to the result.
If in this case we will have for instance the stylesheet as
<?xml version="1.0" encoding="UTF-8" ?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform";
version="1.0">
<xsl:template match="/">
[<xsl:value-of select="test/a"/>]
[<xsl:value-of select="test/b"/>]
</xsl:template>
</xsl:stylesheet>
and the output:
[a]
[b]
Then you can infere that the initial document has test as root element and
at least two children a and b with content "a" and "b" respectivelly.
HTH,
George
-------------------------------------------------------------
George Cristian Bina mailto:george(_at_)oxygenxml(_dot_)com
<oXygen/> XML Editor - http://www.oxygenxml.com/
----- Original Message -----
From: "Bonnefoy Pierre" <pierre(_dot_)bonnefoy(_at_)scania(_dot_)com>
To: <xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com>
Sent: Wednesday, January 07, 2004 11:14 AM
Hi all !
My problem :
An HTML file is generated by an XML file and a stylesheet, serverside.
I've the HTML file and the stylesheet. Can I retrieve the original XML,
and
if yes, how ?
Manythanks,
/Pierre
XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list