RE: [xsl] Name of files being processed

["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: [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.