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Re: How to get and display part of XML document in IE5/6?
- From: "Qinghui Zeng" <zengqh at dfki dot de>
- To: <xsl-list at lists dot mulberrytech dot com>
- Date: Tue, 18 Jun 2002 17:07:18 +0200
- Subject: Re: [xsl] How to get and display part of XML document in IE5/6?
- Organization: DFKI Saarbruecken GmbH, D 66123 Saarbruecken
- References: <000701c2152b$eea51b10$a5bb6086@dfki.unisb.de> <31918.1024247079@www63.gmx.net>
- Reply-to: xsl-list at lists dot mulberrytech dot com
> > " english.xml#xpointer(/document/language/page(id=1)) "
>
> To make it short:
>
> <xsl:value-of
> select="document('english.xml')/document/language/page[@ID=1]"/>
>
That's fine to fetch the part content with this method, but still
a problem is how to pass the parameter(e.g. ID=1) to stylesheet.
In detail, I have five xml files, each for one language(English,
German,...),
and each includes some pages.
---------------------------------
<Doc lang=EN>
<Pages>
<Page ID=1>.....</Page>
......
<Page ID=1>.....</Page>
</Pages>
</Doc>
---------------------------------
Assume, Page 1 is displayed in IE, now I want to display Page 5 by a link,
Considering compatibility with future fully xml-supported explorer, (right?)
In page 1(source), I use such a link :
" english.xml#xpointer(/Doc/Pages/Page(ID=5)) "
but I don't know how I can pass "ID=5" to the template in XSL?
Or I need use other link method ?
Thanks.
Qinghui
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