Question

我正在尝试使用JavaFX创建测验应用程序，因为我正在使用来调用问题

Q1.invoke();
Q2.invoke();

这些问题将显示在UI线程上

public void display(McqQuestion mcqQuestion) {
        resourceAsStream  = getClass().getResourceAsStream("/mcqview.fxml");
        fxmlLoader = new FXMLLoader();
        if (executorService==null) executorService =Executors.newSingleThreadExecutor();
        Parent root = null;
        try {
            root = fxmlLoader.load(resourceAsStream);
            Mcqview controller = fxmlLoader.getController();
            controller.setAnswer1(mcqQuestion.getAnswers().get(0));
            //controller class has setters to accept question properties.
            controller.multipleChoiceQuestionType = this;
            this.view.getBorderPane().setCenter(root);
}

显示问题后，我需要等待直到得到答案为止，如果没有得到答案，则应该调用下一个问题。因此，我在display方法内部引入了一个线程来等待超时

submit = executorService.submit(() -> {
             try {
                    TimeUnit.SECONDS.sleep(2000);
                } catch (InterruptedException e) {
                    e.printStackTrace();
                }
             });

            try {
                submit.get(20,TimeUnit.SECONDS);
                System.out.println("waiting finished");
            } catch (InterruptedException e) {
                e.printStackTrace();
            }

由于future.get();是一个阻塞调用，它也阻塞了UI线程，因此如何在不阻塞UI线程的情况下实现这一目标。

Answer 1

请勿为此目的使用单独的线程。这只会使事情变得更困难。 JavaFX提供了无需等待并发问题的等待方式。

在这种情况下，可以使用PauseTransition处理程序从onFinished完成等待。处理来自事件处理程序的用户输入答案。

private static class Question {

    private final String questionText;
    private final String answers[];
    private final int correctAnswerIndex;

    public Question(String questionText, String[] answers, int correctAnswerIndex) {
        if (answers.length != 3) {
            // for simplicity's sake allow only exactly 3 answers
            throw new IllegalArgumentException();
        }
        this.questionText = questionText;
        this.answers = answers;
        this.correctAnswerIndex = correctAnswerIndex;
    }

}

private VBox questionPane;
private Label questionText;
private Button[] answerButtons;
private PauseTransition pauseTransition;
private Question currentQuestion;

private void answer(int index) {
    pauseTransition.stop(); // no longer wait for timeout
    Alert alert = new Alert(Alert.AlertType.INFORMATION);
    alert.setContentText((index == currentQuestion.correctAnswerIndex)
            ? "correct answer"
            : "incorrect answer");

    // show result and exit
    alert.showAndWait();
    Platform.exit();
}

private void ask(Question question) {
    questionText.setText(question.questionText);
    for (int i = 0; i < 3; i++) {
        answerButtons[i].setText(question.answers[i]);
    }
    currentQuestion = question;
    pauseTransition.playFromStart(); // start timeout timer
}

private void timeout() {
    pauseTransition.stop();
    Alert alert = new Alert(Alert.AlertType.INFORMATION);
    alert.setContentText("your time ran out");

    // cannot use showAndWait form animation directly
    Platform.runLater(() -> {
        // show result and exit
        alert.showAndWait();
        Platform.exit();
    });
}

@Override
public void start(Stage stage) {
    pauseTransition = new PauseTransition(Duration.seconds(10));
    pauseTransition.setOnFinished(evt -> timeout());

    questionText = new Label();
    questionText.setWrapText(true);

    questionPane = new VBox(questionText);
    questionPane.setPrefSize(400, 400);
    answerButtons = new Button[3];

    for (int i = 0; i < 3; i++) {
        final int answerIndex = i;
        Button button = new Button();
        button.setOnAction(evt -> answer(answerIndex));
        answerButtons[i] = button;
        questionPane.getChildren().add(button);
    }

    Scene scene = new Scene(questionPane);

    stage.setScene(scene);
    stage.show();

    Question question = new Question(
            "What is the answer to the ultimate question of life, the universe, and everything?",
            new String[]{"Mew", "42", "Peanut butter"},
            1
    );
    ask(question);
}

您可以轻松地以不同的方式实现超时或回答问题的结果，例如通过询问下一个问题或在完成最后一个问题时显示结果。

Answer 2

对于UI更改，您应该使用

Platform.runLater(() -> {

});

对于Thread，您应该使用：

Task<Void> task = new Task<Void>() {
    @Override
    protected Void call() throws Exception {
                return null;
        }
};

并将任务对象传递给

executorService.submit(task)

希望这会有所帮助

等待任务完成而不会阻塞javafx

2 个答案: