I attached two classes to get properties from a jar. You can modify its to
get other kind of files (I did it).
Finally, you can get content from the package and file into the jar.
In the properties case, you can get a property with something like that:
private static final String LDAP_CASA =
PropertyReaderHelper.getProperty("PROVIDER_URL_CASA");
private static final String CASA_CONTEXT_SEARCH_1 =
PropertyReaderHelper.getProperty("CONTEXT_SEARCH_CASA_1");
private static final String CASA_CONTEXT_SEARCH_2 =
PropertyReaderHelper.getProperty("CONTEXT_SEARCH_CASA_2");
private static final String CASA_PRINCIPAL =
PropertyReaderHelper.getProperty("PRINCIAL_CASA");
private static final String CASA_CREDENTIALS =
PropertyReaderHelper.getProperty("CREDENTIALS_CASA");
I hope it helps you.
Regards,
JP
2010/11/2 KARR, DAVID (ATTSI) <[email protected]>
> > -----Original Message-----
> > From: Dennis Sosnoski [mailto:[email protected]]
> > Sent: Tuesday, November 02, 2010 3:00 PM
> > To: [email protected]
> > Cc: KARR, DAVID (ATTSI)
> > Subject: Re: How to specify wsdlLocation for wsdl bundled in jar?
> >
> > Hi David,
> >
> > I'd think the way to handle this would be to use the service
> > constructor
> > that takes a WSDL URL. You can find the URL at runtime for a file in
> > the
> > classpath by using the classloader getResource(String name) method.
>
> Ok, good. That's simpler. I've verified this works.
>
> > On 11/03/2010 10:49 AM, KARR, DAVID (ATTSI) wrote:
> > >> -----Original Message-----
> > >> From: KARR, DAVID (ATTSI)
> > >> Sent: Monday, November 01, 2010 12:54 PM
> > >> To: [email protected]
> > >> Subject: How to specify wsdlLocation for wsdl bundled in jar?
> > >>
> > >> I'm trying to produce a "client" jar that includes my generated
> > >>
> > > classes
> > >
> > >> and my wsdl and schema.
> > >>
> > >> At the root of the jar file, where the "com" directory exists, I
> > also
> > >> have a "contract" directory, that has a subdir that specifies a
> > >> contract
> > >> version number. In that directory is my wsdl and schemas.
> > >>
> > >> In my call to "wsdl2java", I tried specifying
> > >> "classpath:contract/<versiondir>/<wsdlfilename>".
> > >>
> > >> When I generate this and then run it in the container, I see this
> in
> > >> the
> > >> console:
> > >>
> > >> "Can not initialize the default wsdl from
> > classpath:contract/..."
> > >>
> > >> What should the "wsdlLocation" value be? Do I just remove
> > >>
> > > "classpath:"
> > >
> > >> from it? I thought it had to be a URL, so that seems unlikely.
> > >>
> > > I've gotten past this for now, but there's at least one issue here
> > that
> > > is somewhat mystifying. From what I've read about the idea of
> having
> > a
> > > classpath-based URLStreamHandler, the two options are installing the
> > > handler into the JVM or calling the overloaded URL constructor,
> along
> > > with the instance of the overloaded stream handler. I'm not willing
> > to
> > > install it into the JVM at this point, although I might consider
> this
> > > later (perhaps just to avoid the innocuous "Can not initialize the
> > > default wsdl ..." message at load time).
> > >
> > > I now have a static block in a utility class that creates the
> Service
> > > subclass instance, and it uses the URL constructor that takes a URL,
> > > String, and URLStreamHandler. I pass "classpath:..." as the second
> > > parameter and an instance of my "ClasspathURLStreamHandler" which
> > gets
> > > the system classloader as a constructor argument.
> > >
> > > My ClasspathURLStreamHandler is just this:
> > > ------------------
> > > public class ClasspathURLStreamHandler extends URLStreamHandler
> {
> > > private final ClassLoader classLoader;
> > >
> > > public ClasspathURLStreamHandler(ClassLoader classLoader) {
> > > this.classLoader = classLoader;
> > > }
> > >
> > > @Override
> > > protected URLConnection openConnection(URL url) throws
> > > IOException {
> > > final URL resourceUrl =
> > > classLoader.getResource(url.getPath());
> > > return resourceUrl.openConnection();
> > > }
> > > }
> > > ------------------
> > >
> > > It now apparently finds my WSDL, because I can get the porttype and
> > call
> > > the operation.
> > >
> > > However, what's sort of odd is that it apparently never gets into
> the
> > > "openConnection()" method. So, it seems like this isn't doing
> > anything.
> > >
> > > On the other hand, I tried changing to use the "URL(String spec)"
> > > constructor, but that fails with a MalformedUrlException.
> > >
> > > So, somehow by specifying a URLStreamHandler, even though it's not
> > > actually using it, it causes some sort of default behavior that does
> > > exactly what I'm trying to do, which is looking up the "spec" in the
> > > classpath.
> > >
> > >
>
