8

我使用 guava 的 EventBus,不幸的是它捕获并记录了事件处理程序抛出 RuntimeException 时发生的 InvocationTargetException。我可以禁用此行为吗?

4

3 回答 3

8

就目前而言,这是一个深思熟虑的决定,并在 EventBus 文档中进行了讨论:

一般来说,处理程序不应该抛出。如果他们这样做了,EventBus 将捕获并记录异常。这很少是错误处理的正确解决方案,不应依赖;它仅用于帮助在开发过程中发现问题。

正在考虑替代解决方案,尽管我严重怀疑它们是否会进入第 12 版。

于 2012-02-10T20:00:18.920 回答
4

这是懒惰的代码

public class Events
{
    public static EventBus createWithExceptionDispatch()
    {
        final EventBus bus;

        MySubscriberExceptionHandler exceptionHandler = new MySubscriberExceptionHandler();
        bus = new EventBus(exceptionHandler);
        exceptionHandler.setBus(bus);
        return bus;
    }

    private static class MySubscriberExceptionHandler implements SubscriberExceptionHandler
    {
        @Setter
        EventBus bus;

        @Override
        public void handleException(Throwable exception, SubscriberExceptionContext context)
        {
            ExceptionEvent event = new ExceptionEvent(exception, context);
            bus.post(event);
        }
    }
}

现在,您可以订阅ExceptionEvent.

这是我ExceptionEvent的复制和粘贴

@Data
@Accessors(chain = true)
public class ExceptionEvent
{
    private final Throwable exception;
    private final SubscriberExceptionContext context;
    private final Object extra;

    public ExceptionEvent(Throwable exception)
    {
        this(exception, null);
    }

    public ExceptionEvent(Throwable exception, Object extra)
    {
        this(exception,null,extra);
    }

    public ExceptionEvent(Throwable exception, SubscriberExceptionContext context)
    {
        this(exception,context,null);
    }

    public ExceptionEvent(Throwable exception, SubscriberExceptionContext context, Object extra)
    {
        this.exception = exception;
        this.context = context;
        this.extra = extra;
    }
}
于 2014-10-30T02:28:37.670 回答
0

只需继承 guava EventBus,并编写您自己的自定义事件总线。提示:该类应写在 com.google.common.eventbus 包中,以便覆盖内部方法。

package com.google.common.eventbus;

import com.google.common.util.concurrent.MoreExecutors;

public class CustomEventBus extends EventBus {

    /**
     * Creates a new EventBus with the given {@code identifier}.
     *
     * @param identifier a brief name for this bus, for logging purposes. Should be a valid Java
     *     identifier.
     */
    public CustomEventBus(String identifier) {
        super(
            identifier,
            MoreExecutors.directExecutor(),
            Dispatcher.perThreadDispatchQueue(),
            LoggingHandler.INSTANCE);
    }

    /**
     * Creates a new EventBus with the given {@link SubscriberExceptionHandler}.
     *
     * @param exceptionHandler Handler for subscriber exceptions.
     * @since 16.0
     */
    public CustomEventBus(SubscriberExceptionHandler exceptionHandler) {
        super(
            "default",
            MoreExecutors.directExecutor(),
            Dispatcher.perThreadDispatchQueue(),
            exceptionHandler);
    }

    @Override
    void handleSubscriberException(Throwable e, SubscriberExceptionContext context) {
        throw new EventHandleException(e);
    }
}
于 2016-11-29T03:24:30.757 回答