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Re: get DOM Tree from transformation

From: "VK" <schools_ring@-----.--->
To: NULL
Date: 1/3/2007 12:43:00 PM

Martin Honnen wrote:
> That does not work if document.body is the body of an HTML DOM document.
> The XML DOM implementation (of MSXML) and the HTML DOM implementation
> (of IE/MSHTML) are separate implementations, you are not able to move
> nodes from an XML DOM document to a HTML DOM document.

That's a separate issue so a separate answer. I'm not talking about
"move nodes from an XML DOM document to a HTML DOM document". All I
want is to move nodes from one HTML document to another HTML document.
Don't forget that while both XSL template and XML data are XML
documents, the transformation result is whatever you wanted it to be:
XML, HTML, VML, SVG, plain text etc. It even doesn't have to be
well-formed, not talking about being valid.

my XSL test template is

<?xml version="1.0" encoding="ISO-8859-1"?>
<xsl:stylesheet
 version="1.0"
 xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output
 omit-xml-declaration="yes"
 method="html"
 encoding="ISO-8859-1"
 media-type="text/html" />
<xsl:template match="/">
 <ol>
  <xsl:for-each select="data/item">
   <li><xsl:value-of select="word"/></li>
  </xsl:for-each>
 </ol>
</xsl:template>
</xsl:stylesheet>

my XML test data is

<?xml version="1.0" encoding="ISO-8859-1"?>
<data>
 <item>
  <word>One</word>
 </item>
 <item>
  <word>Two</word>
 </item>
 <item>
  <word>Three</word>
 </item>
</data>

which results after transformation into _HTML_ fragment

<ol>
<li>One</li>
<li>Two</li>
<li>Three</li>
</ol>

There is not a smell of XML in the transformation result and most of
UA's do agree with it - except IE that respects <output> processing
instruction during processing but acts like an idiot right after that.

Sorry, I did not mean to make this thread as a "Microsoft criticism" of
any kind. I'm just upset that among dozens and dozens of MSXML methods
there is not a single one for such trivia.



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