RE: XMLFilter in a URIResolver

Create a SAXSource, whose InputSource is the actual XML file, and whose
XMLReader is the XMLFilter. Return the SAXSource as the result of the
URIResolver.

(untested)

Michael Kay

> -----Original Message-----
> From: owner-xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com 
> [mailto:owner-xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com] On Behalf Of 
> Andrew Welch
> Sent: 11 September 2003 12:13
> To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
> Subject: [xsl] XMLFilter in a URIResolver
>
>
>
> Hi all,
>
> I want to incorporate an XMLFilter into my uri resolver, so 
> that any xml referenced by the uri resolver goes through the 
> XMLFilter.
>
> In my URIResolver I currently have:
>
> if (foundHref.endsWith(".xml")) {
>   return new StreamSource(is, u.toString());                    
> }
>
> where 'is' is the xml InputStream and 'u' is the uri.  This 
> returns an unmodified xml.
>
> What I would like to do is return the result of the xml 
> having gone through my XMLFilter.  Im struggling to find the 
> objects involved, as there is no transformation going on 
> (which is what Im used to).
>
> How can I incorporate my XMLFilter into the return statement?  
>
> Thanks for any help
>
> cheers
> andrew
>
>  XSL-List info and archive:  http://www.mulberrytech.com/xsl/xsl-list

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