Thanks Peter, for opening a new line of inquiry. The solution I came up
with is not unlike Michael's recommendation as forwarded by DaveP.
What I did:
Got real path (in class that initiates transformation):
ServletContext ctx = servlet.getServletContext();
String cssFileName = ctx.getRealPath("/css/nrd.css");
Passed it to the Transformer:
trans.setParameter("cssFile", cssFileName);
Then added param at top of stylesheet, and used it in place:
<xsl:param name="cssFile" select="'passedIn'"/>
<link rel="stylesheet" type="text/css" href="{$cssFile}"/>
Thanks to all for helpful pointers.
Best,
William BC Crandall
bc.crandall [around] earthlink.net
----- Original Message -----
From: <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: 12 July 2004 1:18 AM
Subject: RE: Finding CSS files in webapp
> Xalan or Saxon have no 'base' from which to determine
> the context for the css file, so you'll have to use
> an absolute file reference from the stylesheet.
>
> Came up on the Saxon list this week.
> Its for the document() function, but the same logic applies.
>
> HTH DaveP
>
> <quote>
> > It appears that I must use a full path when using the
> > xsl:result-document command from a saxon servlet.
> >
> > For example: <xsl:result-document href="file:/c:/mywebdir/myfile.htm"
> > format="web">
> >
> > rather than: <xsl:result-document href="file:/../../myfile.htm"
> > format="web">
> >
> > Is this true, or is it possible to use a relative path?
>
> Although it's possible in principle to use a relative path, this isn't
> likely to be convenient in a servlet environment. Any relative URI is
> resolved relative to the base URI of the principal output document, which
in
> the servlet case is going to the servlet output streat, which doesn't have
a
> meaningful URI. You could, when you allocate the principal output
> destination, give it an arbitrary URI for this purpose:
>
> ServletOutputStream out = res.getOutputStream(); StreamResult tout = new
> StreamResult(out); tout.setSystemId("file:/c:/some/uri/");
> transformer.transform(new StreamSource(sourceFile), tout);
>
> However, I'm wondering what you're actually trying to do here. Where do
you
> want to put the extra output files? Is it a location associated with the
> individual end user, or a location associated with the source document, or
> what? Are you trying to split the source document into multiple result
> documents the first time it is accessed, and cache the results for later
> use? Do you actually want to serialize the multiple output documents to
> disk, or do you really want them in (application or session) memory?
>
> > Also, there is no
> > 'c:' on Unix\Linux, so would I use something like
> > 'href=file://mywebdir/myfile.html' ??
>
> I think the correct syntax is file:///usr/dir/file.xml to reference the
UNIX
> file /user/dir/file.xml
> >
> > I'm new to the Servlet environment.. I'm not sure how to obtain the
> > full path if I'm running my Servlet on a hosted machine. Do I have to
> > ask the host provider and hard code the path or is there a dynamic way
> > to do this?
> >
>
> If you're running on a hosted machine then you'll have to write the files
to
> a directory where the servlet has write permission. A Java call such as
>
> getServletContext().getRealPath(source);
>
> will give you path names to files within the directory structure visible
to
> users via a browser - it's not clear whether you want the XML files you
> generate to be directly accessible to users or not.
>
> Michael Kay
> </quote>
>
> --
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