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RE: Re:

Subject: RE: Re:
From: "Andrew Welch" <AWelch@xxxxxxxxxxxxxxx>
Date: Wed, 7 Jan 2004 10:56:11 -0000
RE:  Re:
> 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.

Well if you are going to do that you may as well do:

<xsl:template match="/">
  <xsl:copy-of select="."/>
</xsl:template>

But I think the poster was hoping that you could somehow reverse
engineer the output of a transform with the stylesheet, and obtain some
sort of input xml - which of course, you can't.

Cheers
andrew

 XSL-List info and archive:  http://www.mulberrytech.com/xsl/xsl-list


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