scala - 为什么会调用监管策略默认案例？

Question

嗨，我已经对 Parent actor 应用了监督，而孩子是 Throwing MongoTimeOutException（我故意提出该异常以进行测试），但在监督中调用了默认情况。

这是代码：

class child extends Actor {
 def receive = {
  case insertData(data) => 
     //insertion code ,MongoTimeOutException is raised i have killed the service intentionally 
 }
}

class parent extends Actor {

val child=context.actorOf.....//actor creation code

override val supervisorStrategy = OneForOneStrategy(
                                    maxNrOfRetries = 10, withinTimeRange = 10 seconds) {
    case _:MongoException=>
      log.error("Got some MongosException, Supervision Strategy says Resume")  
      Resume
    case _:ElasticsearchTimeoutException=>
      log.error("Got some ElasticsearchTimeoutException, Supervision Strategy says Resume")  
      Resume
    case _:ElasticsearchException=>
      log.error("Got some ElasticsearchException, Supervision Strategy says Resume")  
      Resume
    case _ => 
      log.error("Got some Exception, Supervision Strategy says Resume")
      Resume
  }

 def receive = {
  case Msg(data) => 
     child ! insertData(data)
 }
}

object test extends App
{
 val parent = system.actorof....//actor creation code
 parent !Msg(data)
}

当我运行此代码时，确实会发生 MongoTimeOutException，但这种情况在监督中被调用。

case _ => 
      log.error("Got some Exception, Supervision Strategy says Resume")
      Resume

为什么呢？据我说，它应该援引这种情况：

case _:MongoException=>
  log.error("Got some MongosException, Supervision Strategy says Resume")  

难道我做错了什么？

请帮我 。

Answer 1

打印默认案例值的类，看看什么是真正的异常，以便您处理它。

这是你的代码

case _ => 
      log.error("Got some Exception, Supervision Strategy says Resume")
      Resume

这是你必须做的（打印异常类）

case ex =>
     log.error(s"Exception $ex of type ${ex.getClass}.Got some Exception, Supervision Strategy says Resume", ex)
    Resume

上面的代码将帮助您了解无法捕获的确切消息是什么。