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Posted to dev@xalan.apache.org by Alessandro Marcellini <am...@tiscalinet.it> on 2000/12/18 11:49:36 UTC

Absolute path is a valid URI?

Hi,
I'm working with Xalan-j_2_0_D01, on Windows NT with sp6, sun jdk1.2.2.
I've modified the example "SimpleTransform" only to transform my files:
I give as XSL file "c:\\alle\\html.xsl", that exists, but I get the
exception:
org.xml.sax.SAXParseException: File c:\alle\html.xsl" not found.
This is my code: Templates templates = processor.process(new
InputSource("c:\\alle\\html.xsl"));

But, if  I give the path in this form: "\\alle\\html.xsl" all works
correctly.
Where's the mistake?
Thanks Alessandro

Re: Absolute path is a valid URI?

Posted by Alessandro Marcellini <am...@tiscalinet.it>.


Ben Sifuentes wrote:

> If you wish to use absolute path do the following:
>
> file:///c:/alle/html.xsl
>
> Good luck!!
> -Ben

That works!!!
Thanks
         Alessandro

RE: Absolute path is a valid URI?

Posted by Ben Sifuentes <bs...@cysive.com>.

If you wish to use absolute path do the following:

file:///c:/alle/html.xsl

Good luck!!
-Ben

-----Original Message-----
From: Alessandro Marcellini [mailto:amarcell@tiscalinet.it]
Sent: Monday, December 18, 2000 5:50 AM
To: xalan-dev@xml.apache.org
Subject: Absolute path is a valid URI?


Hi,
I'm working with Xalan-j_2_0_D01, on Windows NT with sp6, sun jdk1.2.2.
I've modified the example "SimpleTransform" only to transform my files:
I give as XSL file "c:\\alle\\html.xsl", that exists, but I get the
exception:
org.xml.sax.SAXParseException: File c:\alle\html.xsl" not found.
This is my code: Templates templates = processor.process(new
InputSource("c:\\alle\\html.xsl"));

But, if  I give the path in this form: "\\alle\\html.xsl" all works
correctly.
Where's the mistake?
Thanks Alessandro