RE: Name of files being processed

Yes, i'll go with the parameters, then.

Thanks.

Quoting Michael Kay <mhk@xxxxxxxxx>:

>
> 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:  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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