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Re: [xsl] relative URI in href

2008-08-24 12:21:56
-revised 2-
Ok, Florent that makes sense, however, it works only by an absolute path
like

<xsl:include href="/p/package/includee.xsl"/>

while

<xsl:include href="includee.xsl"/> leads to the error message

Had IO Exception with stylesheet file: includee.xsl

includee.xsl resides in the same dir /p/package as the includer.xsl.

String path=package.Main.class.getResource("includer.xsl").getPath();
Source xslt = new StreamSource(new File(path));
xslt.setSystemId(path);
Transformer transformer =
  TransformerFactory.newInstance().newTransformer(xslt);
transformer.transform(srcXml, fo);

Do you spot something wrong here?

regards, Rolf


Florent Georges wrote:
Vladimir Nesterovsky wrote:

  Hi

  
if source refer to a file system you better create it as
    

  
      source = new StreamSource(new File(stylesheet));
or 
      source = new StreamSource(stylesheet);
    

  Or if the stylesheet is got from a byte stream or a SAX events
stream, simply use Source.setSystemId.

  The idea is that in order to resolve URIs, the processor has to know
where the stylesheet is.  If it is got from a File or a URI, it can
know that info, but not otherwise, so you would have to help it.

  Regards,

--drkm






















      
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