If you use a jsp portlet, in your jsp file, you would have something like
this:
StreamSource xmlWeb = new StreamSource(cgiWebUrl);
StreamSource xslResponseItems = new
StreamSource(xslPath+"/"+lang+"/mapSearch.xsl");
TransformerFactory tFactory = TransformerFactory.newInstance();
Transformer transformerItems = tFactory.newTransformer(xslResponseItems);
// here is how to pass the parameter id
transformerItems.setParameter("id",myid);
transformerItems.transform( xmlWeb , new StreamResult(out));
in the xsl stylesheet:
<xsl:param name="id" select="'default value'"/>
Hope it help
Antoine
> -----Message d'origine-----
> De : Renaud Liehn [mailto:[EMAIL PROTECTED]
> Envoy� : jeudi 5 juin 2003 11:23
> � : 'Jetspeed Users List'
> Objet : Help on XSLT
>
>
> Hello,
>
>
> i wish to work with an XML file containing data from articles.
>
> I now How to use An XSLT to show the list with appropriate
> html formating..
>
>
> i search for a way to present only one of this element using
> another XSLT
> but i wish this new XSLT
> to be able to show the detail of one element, the id of this
> element should
> be pass as parameters to the XSLT....
>
> is their a way to do this? does XSLT accept parameters in input ??
>
> Thanks for the help
>
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