Akka classic -发送自定义类消息类型

yfwxisqw  于 2022-11-05  发布在  其他
关注(0)|答案(2)|浏览(173)

我尝试遵循在Akka演员之间发送消息的最佳实践,并阅读以下内容:
https://github.com/akka/akka/blob/v2.6.15/akka-docs/src/test/java/jdocs/actor/ActorDocTest.java#L101-L129
https://doc.akka.io/docs/akka/current/actors.html
最好的做法似乎是定义一个包含要发送的消息的静态类。
通过阅读文档,我定义了以下内容:

import akka.event.Logging;
import akka.event.LoggingAdapter;
import akka.persistence.AbstractPersistentActor;

class RecoveryActor extends AbstractPersistentActor {

    public static class Msg1 {}
    public static class Msg2 {}

    private final LoggingAdapter log = Logging.getLogger(getContext().getSystem(), this);

    /**
     * Handler that will be called on recovery.
     * @return
     */
    @Override
    public Receive createReceiveRecover() {
        return receiveBuilder()
                .match(Msg1.class, this::receiveMsg1)
                .match(Msg2.class, this::receiveMsg2)
                .build();
    }

    private void receiveMsg1(Msg1 msg) {
        log.info("in receive message 1");
    }

    private void receiveMsg2(Msg2 msg) {
        log.info("in receive message 2");
    }

    /**
     * The normal receive method.
     * @return
     */
    @Override
    public Receive createReceive() {
        return receiveBuilder()
                .match(String.class, s -> s.equals("cmd"), s -> persist("evt", this::handleEvent))
                .build();
    }

    private void handleEvent(String event) {
        System.out.println("in handle event");
    }

    @Override
    public String persistenceId() {
        return "simple-accountant"; //best practice, the id should be unique.
    }

}

但我不确定如何使用Msg1Msg2发送消息
我定义了一个新类来发送消息:

import akka.actor.ActorRef;
import akka.actor.ActorSystem;
import akka.actor.Props;

public class ActorMessaging {

    public static void main(String args[]){
        final ActorSystem system = ActorSystem.create("example");
        final ActorRef actor = system.actorOf(Props.create(RecoveryActor.class));

        actor.tell(RecoveryActor.Msg1);

    }
}

但它失败,并出现编译器错误:

java: cannot find symbol
  symbol:   variable Msg1
  location: class recoverydemo.RecoveryActor

如何向类型为Msg1的执行元发送消息?
这是在参与者之间发送消息的最佳实践吗?将消息 Package 在自定义类中?

qhhrdooz

qhhrdooz1#

actor.tell正在接受对象。RecoveryActor.Msg1不是这样的,它甚至不是类引用。尝试发送Msg1的示例如何,如下所示:

actor.tell(new RecoveryActor.Msg1(), actorRef)

为了与远程参与者“对话”,首先需要获得对它的引用。

oalqel3c

oalqel3c2#

我犯了一个愚蠢的错误。我覆盖了不正确的方法。我应该覆盖createReceive

public Receive createReceive() {

        return receiveBuilder()
                .match(RecoveryActor.Msg1.class, this::receiveMsg1)
                .match(Msg2.class, this::receiveMsg2)
                .build();

//        return receiveBuilder()
//                .match(String.class, s -> s.equals("cmd"), s -> persist("evt", this::handleEvent))
//                .build();
    }

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