Question

如何在java中多次调用launch（）我被赋予异常＆＃34; ERROR IN MAIN：java.lang.IllegalStateException：不能多次调用应用程序启动＆＃34;

我在我的java应用程序中创建了rest cleint，当请求到来时，调用javafx并在完成webview operarion后使用Platform.exit（）方法关闭javafx窗口时打开webview。当第二个请求收到此错误时如何重新发送此错误。

JavaFx应用程序代码：

public class AppWebview extends Application  {

    public static Stage stage;

    @Override
    public void start(Stage _stage) throws Exception {

        stage = _stage;
        StackPane root = new StackPane();

        WebView view = new WebView();

        WebEngine engine = view.getEngine();
        engine.load(PaymentServerRestAPI.BROWSER_URL);
        root.getChildren().add(view);
        engine.setJavaScriptEnabled(true);
        Scene scene = new Scene(root, 800, 600);
        stage.setScene(scene);

        engine.setOnResized(new EventHandler<WebEvent<Rectangle2D>>() {
            public void handle(WebEvent<Rectangle2D> ev) {
                Rectangle2D r = ev.getData();
                stage.setWidth(r.getWidth());
                stage.setHeight(r.getHeight());
            }
        });

        JSObject window = (JSObject) engine.executeScript("window");
        window.setMember("app", new BrowserApp());

        stage.show();

    }

    public static void main(String[] args) {
        launch(args);
    }

RestClient方法：调用JavaFX应用程序

// method 1 to lanch javafx
javafx.application.Application.launch(AppWebview.class);

// method 2 to lanch javafx
String[] arguments = new String[] {"123"};
AppWebview .main(arguments);

Answer 1

您不能多次在JavaFX应用程序上调用launch()，这是不允许的。

来自javadoc：

It must not be called more than once or an exception will be thrown.

建议定期显示一个窗口

只需拨打Application.launch()一次即可。
使用Platform.setImplicitExit(false)在后台运行JavaFX运行时，以便在隐藏最后一个应用程序窗口时JavaFX不会自动关闭。
下次需要另一个窗口时，请在show()中包含窗口Platform.runLater()，以便在JavaFX应用程序线程上执行调用。

如果您要混合使用Swing，则可以使用JFXPanel代替Application，但使用模式与上述类似。

Wumpus示例

import javafx.animation.PauseTransition;
import javafx.application.*;
import javafx.geometry.Insets;
import javafx.scene.Scene;
import javafx.scene.control.Label;
import javafx.stage.Stage;
import javafx.util.Duration;

import java.util.*;

// hunt the Wumpus....
public class Wumpus extends Application {
    private static final Insets SAFETY_ZONE = new Insets(10);
    private Label cowerInFear = new Label();
    private Stage mainStage;

    @Override
    public void start(final Stage stage) {
        // wumpus rulez
        mainStage = stage;
        mainStage.setAlwaysOnTop(true);

        // the wumpus doesn't leave when the last stage is hidden.
        Platform.setImplicitExit(false);

        // the savage Wumpus will attack
        // in the background when we least expect
        // (at regular intervals ;-).
        Timer timer = new Timer();
        timer.schedule(new WumpusAttack(), 0, 5_000);

        // every time we cower in fear
        // from the last savage attack
        // the wumpus will hide two seconds later.
        cowerInFear.setPadding(SAFETY_ZONE);
        cowerInFear.textProperty().addListener((observable, oldValue, newValue) -> {
            PauseTransition pause = new PauseTransition(
                    Duration.seconds(2)
            );
            pause.setOnFinished(event -> stage.hide());
            pause.play();
        });

        // when we just can't take it  anymore,
        // a simple click will quiet the Wumpus,
        // but you have to be quick...
        cowerInFear.setOnMouseClicked(event -> {
            timer.cancel();
            Platform.exit();
        });

        stage.setScene(new Scene(cowerInFear));
    }

    // it's so scary...
    public class WumpusAttack extends TimerTask {
        private String[] attacks = {
                "hugs you",
                "reads you a bedtime story",
                "sings you a lullaby",
                "puts you to sleep"
        };

        // the restaurant at the end of the universe.
        private Random random = new Random(42);

        @Override
        public void run() {
            // use runlater when we mess with the scene graph,
            // so we don't cross the streams, as that would be bad.
            Platform.runLater(() -> {
                cowerInFear.setText("The Wumpus " + nextAttack() + "!");
                mainStage.sizeToScene();
                mainStage.show();
            });
        }

        private String nextAttack() {
            return attacks[random.nextInt(attacks.length)];
        }
    }

    public static void main(String[] args) {
        launch(args);
    }
}

Answer 2

试试这个，我试过这个并找到了成功的

@Override
public void start() {
    super.start();
    try {
                    // Because we need to init the JavaFX toolkit - which usually Application.launch does
                    // I'm not sure if this way of launching has any effect on anything
        new JFXPanel();

        Platform.runLater(new Runnable() {
            @Override
            public void run() {
                // Your class that extends Application
                new ArtisanArmourerInterface().start(new Stage());
            }
        });
    } catch (Exception e) {
        e.printStackTrace();
    }
}

Answer 3

我使用类似其他答案的东西。

private static volatile boolean javaFxLaunched = false;

public static void myLaunch(Class<? extends Application> applicationClass) {
    if (!javaFxLaunched) { // First time
        Platform.setImplicitExit(false);
        new Thread(()->Application.launch(applicationClass)).start();
        javaFxLaunched = true;
    } else { // Next times
        Platform.runLater(()->{
            try {
                Application application = applicationClass.newInstance();
                Stage primaryStage = new Stage();
                application.start(primaryStage);
            } catch (Exception e) {
                e.printStackTrace();
            }
        });
    }
}

如何在java中多次调用launch（）

3 个答案: