Subject: RE: [xsl] Name of files being processed From: "Michael Kay" <mhk@xxxxxxxxx> Date: Thu, 29 Jul 2004 13:54:24 +0100 |
The form document('relative.uri', node) allows you to select a document relative to the base URI of a node in the source document, without actually knowing the base URI. In 2.0 you can obtain the base URI of any node using the base-uri() function. But in 1.0 the only way to get it is to pass it in as a parameter. Michael Kay > -----Original Message----- > From: xptm [mailto:xptm@xxxxxxx] > Sent: 29 July 2004 13:17 > To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx > Subject: [xsl] Name of files being processed > > Hi: > > I'm making a transformation from Java like this: > > BufferedReader br = new BufferedReader(new > InputStreamReader(new > FileInputStream("INPUT.xml"))); > PrintWriter out = new PrintWriter(new > FileOutputStream("OUTPUT.xml")); > try { > TransformerFactory xformFactory = > TransformerFactory.newInstance(); > Source xsl = new StreamSource("Testes12.xsl"); > Transformer stylesheet = xformFactory.newTransformer(xsl); > Source request = new StreamSource(br); > Result response = new StreamResult(out); > stylesheet.transform(request, response); > > in other words, i'm transforming the file named INPUT into the file > named OUTPUT. This names can, of course, be variable. > > How can i, inside my XSLT, know the names of the files been > processed? I > want to know this because i want to use document() with a file name > related with the one been processed. > > In my Java example maybe i can just use > > stylesheet.setParameter("inputfilename", ...); > > but that's not a ideal solution for me. > > Thxs.
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