This is the mail archive of the xsl-list@mulberrytech.com mailing list .


Index Nav: [Date Index] [Subject Index] [Author Index] [Thread Index]
Message Nav: [Date Prev] [Date Next] [Thread Prev] [Thread Next]
Other format: [Raw text]

Re: How to get and display part of XML document in IE5/6?


Hello Quinghui,

in what framework do you want to use XInclude and XPointer? I think it's not
possible to do it client-side. At least IE has no XInclude-transformer. And
I don't know any browser which has one, but maybe Mozilla has still more
secrets.

On the other hand a two-step transformation is also not possible I think. 
But where is the problem to do it in one step/transformation process?

<xsl:template match="/">
  <xsl:apply-templates select="id('ebnf')"/>
</xsl:template>

<xsl:template match="the-element-which-has-ebnf-as-id">
  <!-- transform it to what you want -->
  <html>
    <head><title>test</title></head>
    <body>
      This is the element which has 'ebnf' as ID.
    </body>
  </html>
</xsl:template>

Regards,

Joerg
 
> I want to get and display part of a linked XML document, I can make
> use of XLink and XPointer to refer to the target, e.g.
> http://www.w3.org/TR/1998/REC-xml-19980210.xml#xpointer(id("ebnf"))
> but how can I display the content in normal explorer(e.g. IE5/IE6)?
> I have not found the solutions or samples to solve this problem.
> 
> In another way, I try to write stylesheet to extract the target to be a 
> separate document. then I can display it in explorer with another 
> stylesheet.  but XSL does not provide such function to extract the
> part of XML document, without stripping the tag.
> 
> Can you give me some suggestion?
> 
> Thanks 
> 
> Qinghui

-- 
GMX - Die Kommunikationsplattform im Internet.
http://www.gmx.net


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


Index Nav: [Date Index] [Subject Index] [Author Index] [Thread Index]
Message Nav: [Date Prev] [Date Next] [Thread Prev] [Thread Next]