That is why the bundle activator creates a bundle-singleton of itself,
that way the app can access the OSGi world. In my case to register
itself as a service.
@Override
public void start(Stage primaryStage) throws Exception {
....
primaryStage.show();
Dictionary<String, ?> properties = createDictionary();
BundleContext bundleContext =
UdooActivator.bundleActivator().getBundleContext();
bundleContext.registerService(com.cuhka.home.application.Application.class,
this, properties);
}
Maurice.
Op 20-02-16 om 15:08 schreef Stephen Winnall:
Hi Maurice
I have done something similar, but it has the following drawback in my view:
the class launched (Udoo15App in your case) does not run under OSGi control, so
it has no access to OSGi bundles or services, nor is it accessible by them. If
you don’t need that, you're OK. But I need that class to be part of the OSGi
world because other bundles/services are going to add parts to the UI as they
are instantiated.
Steve
On 20 Feb 2016, at 14:33, Maurice <i...@cuhka.com> wrote:
For my OSGi based JavaFX solution on the Udoo Quad (ARM based Linux) I created
a service that publishes the application in the context.The application does as
little as possible. It sets up the primary stage as fullscreen and puts a
stackpane in it. Initially the stackpane displays a 'boot logo', until the
actual desktop bundle is started and registered with the application. Note that
you have to start the application on a separate thread, as the thread will be
blocked.
On Java 8 this means that although the application bundle can't be updated in a
running OSGi container, but that is why the desktop exists. On startup it
registers itself, and thus the application content, with the application, and
when it is stopped it removes the content from the application. The application
has thus rarely to be updated itself.
Regards,
Maurice.
public class UdooActivator implements BundleActivator {
private static UdooActivator activator;
private BundleContext context;
static UdooActivator bundleActivator() {
return requireNonNull(activator, "activator not set");
}
@Override
public void start(BundleContext context) throws Exception {
this.context = context;
activator = this;
new Thread(() -> Application.launch(Udoo15App.class), "JavaFX Desktop
launcher").start();
}
@Override
public void stop(BundleContext context) throws Exception {
Platform.exit();
}
public BundleContext getBundleContext() {
return context;
}
}
Op 20-02-16 om 01:28 schreef Stephen Winnall:
Anirvan, Kevin
Thanks for this.
I’m an expert neither in JavaFX nor in OSGi, but I think the basis of the
JavaFX/OSGi incompatibility is control. To work with OSGi, JavaFX has to
relinquish control of its startup sequence to OSGi in such a way that
javafx.application.Application (or its proxy) is instantiated by OSGi and
submits to OSGi’s bundle/service lifecycle. AN OSGi expert can probably
formulate this better…
Platform.startup(runnable) /might/ do it. Platform.launch(class) doesn’t
because the object thereby instantiated is always under the control of JavaFX -
and thus not of OSGi.
I’m not comfortable using JFXPanel: if I wanted to use Swing I wouldn’t be
trying to use JavaFX. But thank you for the hint.
Steve
On 19 Feb 2016, at 16:41, Kevin Rushforth<kevin.rushfo...@oracle.com> wrote:
And for JDK 9 there is now:
Platform.startup(Runnable);
-- Kevin
Anirvan Sarkar wrote:
Hi Stephen,
FYI, there is another way of initializing JavaFX runtime. Just use:
new JFXPanel();
It is documented[1] that FX runtime is initialized when the first JFXPanel
instance is constructed.
Also JavaFX 9 will provide an official API to start the FX platform [2] [3].
[1]
https://docs.oracle.com/javase/8/javafx/api/javafx/application/Platform.html#runLater-java.lang.Runnable
<https://docs.oracle.com/javase/8/javafx/api/javafx/application/Platform.html#runLater-java.lang.Runnable>-
[2]https://bugs.openjdk.java.net/browse/JDK-8090585
<https://bugs.openjdk.java.net/browse/JDK-8090585>
[3]
http://download.java.net/jdk9/jfxdocs/javafx/application/Platform.html#startup-java.lang.Runnable
<http://download.java.net/jdk9/jfxdocs/javafx/application/Platform.html#startup-java.lang.Runnable>-
On 18 February 2016 at 20:08, Stephen Winnall<st...@winnall.ch>
<mailto:st...@winnall.ch> wrote:
As I understand it, there are two ways of activating JavaFX:
1) sub-class javafx.application.Application or
2) call javafx.application.Application.launch()
Op 20-02-16 om 01:28 schreef Stephen Winnall:
Anirvan, Kevin
Thanks for this.
I’m an expert neither in JavaFX nor in OSGi, but I think the basis of the
JavaFX/OSGi incompatibility is control. To work with OSGi, JavaFX has to
relinquish control of its startup sequence to OSGi in such a way that
javafx.application.Application (or its proxy) is instantiated by OSGi and
submits to OSGi’s bundle/service lifecycle. AN OSGi expert can probably
formulate this better…
Platform.startup(runnable) /might/ do it. Platform.launch(class) doesn’t
because the object thereby instantiated is always under the control of JavaFX -
and thus not of OSGi.
I’m not comfortable using JFXPanel: if I wanted to use Swing I wouldn’t be
trying to use JavaFX. But thank you for the hint.
Steve
On 19 Feb 2016, at 16:41, Kevin Rushforth <kevin.rushfo...@oracle.com> wrote:
And for JDK 9 there is now:
Platform.startup(Runnable);
-- Kevin
Anirvan Sarkar wrote:
Hi Stephen,
FYI, there is another way of initializing JavaFX runtime. Just use:
new JFXPanel();
It is documented[1] that FX runtime is initialized when the first JFXPanel
instance is constructed.
Also JavaFX 9 will provide an official API to start the FX platform [2] [3].
[1]
https://docs.oracle.com/javase/8/javafx/api/javafx/application/Platform.html#runLater-java.lang.Runnable
<https://docs.oracle.com/javase/8/javafx/api/javafx/application/Platform.html#runLater-java.lang.Runnable>-
[2] https://bugs.openjdk.java.net/browse/JDK-8090585
<https://bugs.openjdk.java.net/browse/JDK-8090585>
[3]
http://download.java.net/jdk9/jfxdocs/javafx/application/Platform.html#startup-java.lang.Runnable
<http://download.java.net/jdk9/jfxdocs/javafx/application/Platform.html#startup-java.lang.Runnable>-
On 18 February 2016 at 20:08, Stephen Winnall <st...@winnall.ch>
<mailto:st...@winnall.ch> wrote:
As I understand it, there are two ways of activating JavaFX:
1) sub-class javafx.application.Application or
2) call javafx.application.Application.launch()
