我正在尝试使用2个actor作为源,然后是一个合并结,然后是1个接收器来构建和运行akka流(在Java DSL中):
I am trying to build and run an akka stream flow (in Java DSL) with 2 actors as sources, then a merge junction and then 1 sink:
Source<Integer, ActorRef> src1 = Source.actorRef(100, OverflowStrategy.backpressure()); Source<Integer, ActorRef> src2 = Source.actorRef(100, OverflowStrategy.backpressure()); Sink<Integer, BoxedUnit> sink = Flow.of(Integer.class).to(Sink.foreach(System.out::println)); RunnableFlow<BoxedUnit> closed = FlowGraph.factory().closed(sink, (b, out) -> { UniformFanInShape<Integer, Integer> merge = b.graph(Merge.<Integer>create(2)); b.from(src1).via(merge).to(out); b.from(src2).to(merge); }); closed.run(mat);我的问题是如何获取对源actor的ActorRef引用以便向其发送消息?如果有1个actor,我将不会使用图形生成器,然后.run()或runWith()方法将返回ActorRef对象。但是,如果有许多来源参与者,该怎么办?
My question is how do I obtain ActorRef references to the source actors in order to send them messages? In case of 1 actor, I wouldn't be using graph builder, and then the .run() or runWith() method would return the ActorRef object. But what to do in case of many source actors? Is it even possible to materialize such a flow?
推荐答案在有人需要的情况下回答我自己的问题。
Answering my own question in case someone needs it.
使用jrudolph的建议,我可以使用这样的actor(在实际代码中,我做的比2个ActorRefs列表还要好):
Using jrudolph's advice, I was able to use actors like this (in actual code I did something nicer than a list of 2 ActorRefs):
Source<Integer, ActorRef> src1 = Source.actorRef(100, OverflowStrategy.fail()); Source<Integer, ActorRef> src2 = Source.actorRef(100, OverflowStrategy.fail()); Sink<Integer, BoxedUnit> sink = Flow.of(Integer.class).to(Sink.foreach(System.out::println)); RunnableFlow<List<ActorRef>> closed = FlowGraph.factory().closed(src1, src2, (a1, a2) -> Arrays.asList(a1, a2), (b, s1, s2) -> { UniformFanInShape<Integer, Integer> merge = b.graph(Merge.<Integer>create(2)); b.from(s1).via(merge).to(sink); b.from(s2).to(merge); }); List<ActorRef> stream = closed.run(mat); ActorRef a1 = stream.get(0); ActorRef a2 = stream.get(1);更多推荐
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