Question

我正在关注akka.io容错http://doc.akka.io/docs/akka/current/java/fault-tolerance.html中的代码。我已将此代码作为参考。我的要求如下：让我们假设演员在一条消息上崩溃了并由他的主管重新启动。然后他开始处理下一个在他的邮箱中的消息。导致崩溃的消息是 '丢弃'。但我想在特定时间内（假设3次）处理相同的动作，并在它们之间定义一个间隔（假设为1秒）。如何使用akka监督来做到这一点。实际上通过actor我试图检查一个特定的服务api是否正在工作（即给出一些例外）。所以如果特定的尝试有任何异常，（假设404未找到），重新发送消息给失败的工作者直到达到supervisorStrategy指定的maxNrOfRetries。如果工作人员失败了“maxNrOfRetries”次，那么只记录为“此xx消息的最大尝试次数”。如何在java中执行此操作。

我的主管班：

public class Supervisor extends UntypedActor {


 private static SupervisorStrategy strategy =

 new OneForOneStrategy(3, Duration.create("1 minute"),
  new Function<Throwable, Directive>() {
    @Override
    public Directive apply(Throwable t) {
      if (t instanceof Exception) {
        return restart();
      }else if (t instanceof IllegalArgumentException) {
        return stop();
      } else {
        return escalate();
      }
    }
  });

 @Override
 public SupervisorStrategy supervisorStrategy() {
 return strategy;


}
public void onReceive(Object o) {
if (o instanceof Props) {
  getSender().tell(getContext().actorOf((Props) o), getSelf());
} else {
  unhandled(o);
}


 }
}

儿童班：

public class Child extends UntypedActor {


  public void onReceive(Object o) throws Exception {
if (o instanceof String) {
Object response = someFunction( (String) message);//this function returns either successfull messgae as string or exception
if(response instanceOf Exception) {
     throw (Exception) response;
   } 
   else
     getSender().tell(response, getSelf())
}else {
  unhandled(o);
}


}

}

创建演员：

Props superprops = Props.create(Supervisor.class);
ActorRef supervisor = system.actorOf(superprops, "supervisor");
ActorRef child = (ActorRef) Await.result(ask(supervisor,
Props.create(Child.class), 5000), timeout);
child.tell("serVice_url", ActorRef.noSender());

对于service_url，如果发生故障，我想重复该过程。但它没有发生。如果将creatng actor中的下一行写为child.tell("serVice_url_2", ActorRef.noSender());，则此行被引导但我想处理相同的动作（发生故障）一段特定的时间（假设3次），并且它们之间有一个定义的间隔。请指导我实现这一目标。

Answer 1

我想我已经开发了一种方法。虽然我仍然需要在生产水平上进行测试。我正在写下面的答案，因为它可能对尝试实现同样的事情有所帮助。如果有人找到更好的方法然后欢迎他/她。这里要提一下，通过这种方法 Supervisor在一个时间范围内处理相同的动作（发生故障的消息）特定次数（假设3次）。我无法定义间隔他们之间。 这是代码。主管班。

public class MyUntypedActor extends UntypedActor {
//here I have given Max no retrilas as 10.I will controll this number from logic as per my own requirements.But user given number of retrials can not exceed 10.
private static SupervisorStrategy strategy = new AllForOneStrategy(10, Duration.create(5, TimeUnit.MINUTES),
        new Function<Throwable, SupervisorStrategy.Directive>() {
            @Override
            public SupervisorStrategy.Directive apply(Throwable t) {
                if (t instanceof Exception) {
                    //System.out.println("exception" + "*****" + t.getMessage() + "***" + t.getLocalizedMessage());
                    return restart();
                } else if (t instanceof NullPointerException) {
                    return restart();
                } else if (t instanceof IllegalArgumentException) {
                    return stop();
                } else {
                    return escalate();
                }
            }
        });

@Override
public SupervisorStrategy supervisorStrategy() {
    return strategy;
}

public void onReceive(Object o) {
    if (o instanceof Props) {
        getSender().tell(getContext().actorOf((Props) o), getSelf());
    } else {
        unhandled(o);
    }
}
}

我们将编写逻辑的子类。

public class Child extends UntypedActor {
//Through preRestart it will push the message for which exception occured before the restart of the child
@Override
public void preRestart(final Throwable reason, final scala.Option<Object> message) throws Exception {
    System.out.println("reStarting :::" + message.get());
    SetRules.setRemainingTrials(SetRules.remainingTrials + 1);
    getSelf().tell(message.get(), getSender());
};

public void onReceive(Object o) throws Exception {

    if (o instanceof Exception) {
        throw (Exception) o;
    } else if (o instanceof Integer) {
    } else if (o.equals("get")) {
        getSender().tell("get", getSelf());
    } else if (o instanceof String) {

        try {
            // here either we can write our logic directly or for a better
            // approach can call a function where the logic will be excuted.
            getSender().tell("{\"meggase\":\"Succesfull after " + SetRules.remainingTrials + " retrials\"}",
                    getSelf());
        } catch (Exception ex) {
            if (SetRules.remainingTrials == SetRules.noOfRetries) {
                getSender().tell("{\"meggase\":\"Failed to connect after " + SetRules.noOfRetries + " retrials\"}",
                        getSelf());
            } else {
                Exception value1 = ex;
                throw (Exception) value1;
            }
        }
    } else {
        unhandled(o);
    }
}
}

具有关于用户信息的SetRules类提供noOfRetrials，并且还存储有关通过remainingTrials重试的每个重试次数的信息

public class SetRules {

public static int noOfRetries;
public static int remainingTrials;

public SetRules(int noOfRetries, int remainingTrials) {
    super();
    SetRules.noOfRetries = noOfRetries;
    SetRules.remainingTrials = remainingTrials;
}

public int getRemainingTrials() {
    return remainingTrials;
}

public static void setRemainingTrials(int remainingTrials) {
    SetRules.remainingTrials = remainingTrials;
}
}

现在让我们创建一个actor。

Props superprops = Props.create(MyUntypedActor.class);
SetRules setRules=new SetRules(3,0);
ActorSystem system = ActorSystem.create("helloakka");
ActorRef supervisor = system.actorOf(superprops, "supervisor");
ActorRef child = (ActorRef) Await.result(ask(supervisor, Props.create(Child.class), 5000), Duration.create(5, "minutes"));
Future<Object> future = Patterns.ask(child, service_Url, new Timeout(Duration.create(5, "minutes")));
Object result =  Await.result(future, Duration.create(5, "minutes"));
System.out.println(result);

使用Actor监督，如果发生故障，如何以定义的间隔重试相同的消息达定义的次数

1 个答案: