宁愿这是一个非常简单的概念,但是由于我对JavaFX的并发性完全陌生.我一直在努力理解这个概念.

在构建一个非常简单的JavaFX应用程序时,我想在后台执行一些冗长的任务,并使UI免受冻结.

在下面的示例中,我试图创建一个简单的后台任务服务,然后使用ControlsFX Dialog在主UI窗口上显示进度栏.

但是,我想知道如何为该服务提供一些输入参数并从中获取布尔输出.这在C#BackgroundWorker中确实很简单,但是在JavaFX中仍然无法弄清楚.任何提示将是一个很大的帮助.

尝试了这个简单的例子

public void runTask(Stage stage) throws URISyntaxException, IOException {

Service service = new Service() {

@Override

protected Task createTask() {

return new Task() {

@Override

protected Void call() throws InterruptedException, URISyntaxException, IOException {

// some time consuming task here

// use the input arguments and perform some action on it

// then set the process result to a Boolean and return after the task is completed

// also keep hold back any other process to from executing on UI

return null;

}

};

}

};

Dialogs.create()

.owner(stage)

.title("Performing Task!")

.masthead("Please wait...")

.showWorkerProgress(service);

service.setOnSucceeded(new EventHandler() {

@Override

public void handle(WorkerStateEvent event) {

System.out.println("done:" + event.getSource().getValue());

}

});

service.setOnFailed(new EventHandler() {

@Override

public void handle(WorkerStateEvent event) {

throw new UnsupportedOperationException("Failed."); //To change body of generated methods, choose Tools | Templates.

}

});

service.setOnCancelled(new EventHandler() {

@Override

public void handle(WorkerStateEvent event) {

throw new CancellationException("Cancelled."); //To change body of generated methods, choose Tools | Templates.

}

});

service.start();

}