Subject: RE: [xsl] XMLFilter in a URIResolver From: "Michael Kay" <mhk@xxxxxxxxx> Date: Thu, 11 Sep 2003 12:48:26 +0100 |
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@xxxxxxxxxxxxxxxxxxxxxx > [mailto:owner-xsl-list@xxxxxxxxxxxxxxxxxxxxxx] On Behalf Of > Andrew Welch > Sent: 11 September 2003 12:13 > To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx > 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
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