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Re: How to get and display part of XML document in IE5/6?
- From: Joerg dot Heinicke at gmx dot de
- To: xsl-list at lists dot mulberrytech dot com
- Date: Sat, 15 Jun 2002 06:19:52 +0200 (MEST)
- Subject: Re: [xsl] How to get and display part of XML document in IE5/6?
- References: <00ae01c213dd$8c8fa700$a5bb6086@dfki.unisb.de>
- Reply-to: xsl-list at lists dot mulberrytech dot com
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
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