Question

我试图自动运行任务（全部30个）。

为此，我建立了一个单身人士：

public class PortalSingleton {
    private static final Logger LOG = LoggerFactory.getLogger(PortalSingleton.class);
    private static final int INITIAL_DELAY = 0;
    private static final int DELAY = 30;

    private static volatile ScheduledExecutorService instance;
    private static HomepageView homeView = new HomepageView();

    private PortalSingleton() {}

    public static final void refreshGridHomePageAutomatically() {
        Runnable task = () -> UI.getCurrent().access(() -> {
            homeView.refreshGrid();
            LOG.info("The grid has been refreshed Automatically");
        });
        getInstance().scheduleWithFixedDelay(task, INITIAL_DELAY, DELAY, TimeUnit.SECONDS);
    }

    public final static ScheduledExecutorService getInstance() {
        if (instance == null) {
            synchronized (ScheduledExecutorService.class) {
                if (instance == null) {
                    instance = Executors.newScheduledThreadPool(1);
                }
            }
        }
        return instance;
    }
}

但是，我没有任何问题/错误而且我没有我的日志消息，我的网格也没有刷新..

预期的行为是：

我的网格刷新
请参阅log msg

即使我删除了homeView.refreshGrid();行，也没有我的日志信息...

我做错了什么？

谢谢，

编辑：我这样称呼：PortalSingleton.refreshGridHomePageAutomatically();

EDIT2，谢谢@Holger：

public class PortalSingleton {
    private static final Logger LOG = LoggerFactory.getLogger(PortalSingleton.class);
    private static final int INITIAL_DELAY = 0;
    private static final int DELAY = 30;

    private static final ScheduledExecutorService instance = Executors.newScheduledThreadPool(1);
    private static HomepageView homeView = new HomepageView();

    private PortalSingleton() {
    }

    public static final void refreshGridHomePageAutomatically() {
        Runnable task = () -> UI.getCurrent().access(() -> {
            homeView.refreshGrid();
            LOG.info("The grid has been refreshed Automatically");
        });
        try {
            getInstance().scheduleWithFixedDelay(task, INITIAL_DELAY, DELAY, TimeUnit.SECONDS);
        } catch (Exception e) {
            LOG.error("error" + e);
        }
    }

    public final static ScheduledExecutorService getInstance() {
        return instance;
    }
}

Answer 1

安排操作时，如果发生异常，则不会收到反馈。相反，它将停止执行它：

ScheduledExecutorService.scheduleWithFixedDelay(…)：

...如果任务的任何执行遇到异常，则后续执行被禁止。

因此，您必须在操作本身中使用try … catch块来报告，例如在lambda表达式中定义Runnable：

Runnable task = () -> {
    try { UI.getCurrent().access(…); }
    catch (Exception e) { LOG.error("error" + e); }
};

我从一个非UI线程调用UI.getCurrent()看起来很可疑，我怀疑它在尝试调用方法时返回null导致NullPointerException。< / p>

单身人士，无法进入可运行的任务

1 个答案: