RegistryBuilder.constructDefaultRegistry() searches for files named
META-INF/hivemodule.xml only.
To find your services.xml in the classpath try URLResource instead
of FileResource:
private static Registry buildRegistry(String fileName)
{
RegistryBuilder builder = new RegistryBuilder();
ClassResolver resolver = new DefaultClassResolver();
builder.processModules (resolver);
URL url = ThisClass.getClass().getResource(fileName);
if (url == null)
throw new NullPointerException("No resource named '" + file
+ "'.");
builder.processModule(resolver, new URLResource(url));
return builder.constructRegistry(Locale.getDefault());
}
}
For more examples take a look into the source of ExampleUtils class.
Achim
Am Thu, 17 Nov 2005 08:31:44 +0100 schrieb Alex Chew <[EMAIL PROTECTED]>:
Hi, there
How can i build service registry using class resource from my jar file?
i've put my service.xml in jar file, and try
RegistryBuilder.constructDefaultRegistry(),but i could not get service
point.
I can get everything run ok using File Resource, but i must edit java
code
when deploy my WAR file to different directorys.
how can i use just private Registry registry =
buildRegistry("service.xml");//service.xml
will build in jar file
not following lines:
private Registry registry =
buildRegistry("E:/MyProj/main/src/java/service.xml");
private static Registry buildRegistry(String fileName)
{
RegistryBuilder builder = new RegistryBuilder();
ClassResolver resolver = new DefaultClassResolver();
builder.processModules(resolver);
builder.processModule(resolver, new FileResource(fileName));//can i use
Class Resource here?
return builder.constructRegistry(Locale.getDefault());
}
Regards
Alex
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