Che*_*rry 1 scala path actor akka
这是演员代码:
import akka.actor.Actor
class OneActor extends Actor {
def receive = {
//what I should call here to get ?
case _ => println(physicalPaht)
}
}
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我可以使用一些"内置"变量:
有任何想法吗?
更新
也是一个self.path.address但它只返回root actor的路径.
Che*_*rry 10
akka.serialization.Serialization.serializedActorPath(self) 应该使用.
import akka.actor.Actor
class OneActor extends Actor {
def receive = {
case _ => println(akka.serialization.Serialization.serializedActorPath(self))
}
}
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