I've found how to do it :)
I've created a class which implements the javax.xml.transform.URIResolve
In this class, the function resolve(String href, String base) only use the href
parameter in order to find required file as resource using getClass
().getResource(href);
Finally, I return a new StreamSource() with the new URL object created above ..
And it works
Thanks anyway :)
Selon cmalderez(_at_)free(_dot_)fr:
Hello,
I'm using JBoss v3.2.3 with Tomcat 4.1.29.
I've build a small Web application using JSP, EJB producing XML document
(using
JDOM) and displaying data as HTML using XSL stylesheets.
I want to separate some XSL templates into other XSL files and load them from
my primary XSL stylesheet.
Easy using: <xsl:include href="foo.xsl"/>
But all my JSP and XSL files are packaged into a WAR(jar) file and deployed
into JBoss with my EJB package.
The Web Package (WAR file) is mounted using the web-uri: mytool
The Web Package contains 2 folders 'jsp/' with all JSP pages & 'xsl/' with
all
XSL stylesheets.
I access to my JSP using this URL http://localhost:8080/mytool/jsp/myjsp.jsp
In my JSP I can't load my XSL stylesheets as file because they are in a JAR.
So I load them using the getClass().getResource('myxsl.xsl');
But how can I specify the correct URI for the second XSL file 'foo.xsl' which
must be loaded from the 'myxsl.xsl' ?
I can't use <xsl:include href="foo.xsl"/> anymore ...
I've noticed that the following line works <xsl:include
href="http://localhost:8080/mytool/xsl/foo.xsl"/> ... but I can't let hard
configuration like that in XSL files.
How can I get a correct mount point in order to load the 'foo.xsl'
from 'myxsl.xsl' ??
Thanks !
XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list