2016-09-14 98 views
0

首先,我对scala比较陌生。Scala模式匹配总是默认情况下

考虑以下简单的类NullLogService,它将结果记录到控制台。

package services 

import akka.actor._ 
import com.amazonaws.services.simpleemail.model.SendRawEmailResult 
import model.Email 

import scala.util.Try 

object NullLogService { 
    case class Log(email: Email, amazonResult: Try[_]) 
    case class LogError(email: Email, amazonException: Try[_]) 
} 

class NullLogService extends Actor { 
    import NullLogService._ 

    def receive = { 
    case Log(email, amazonResult) => println("Email to: " + email.to + ". AmazonResultSucess:" + amazonResult.isSuccess.toString + ". AmazonResult: " + amazonResult.get.toString) 
    case LogError(email, amazonException) => println("Error: Email to: " + email.to + ". AmazonException:" + amazonException.get.toString) 
    case [email protected]_ => println("Default case: "+default.toString) 
    } 
} 

它总是打印Default case: Log(TO:[email protected],Success({MessageId: Fake_Sent_OK}))

我不知道发生了什么,因为类型相同(Log(Email,Try(_)))!它应该去“登录”的情况下,但总是下降到默认情况下!

让我疯狂。

代码涉及

主类:

import java.io.File 
import java.util.concurrent.TimeUnit 

import akka.actor.SupervisorStrategy.Resume 
import akka.actor._ 
import akka.routing.RoundRobinPool 
import akka.util.Timeout 
import com.thenewmotion.akka.rabbitmq._ 
import com.typesafe.config.{ConfigFactory, ConfigParseOptions} 
import model.Email 
import services.EmailService.EmailInfo 
import services.{EmailService, LogService, NullLogService} 
import utils.StringUtils 

import scala.concurrent.duration.Duration 

object Main extends App { 
    implicit val system = ActorSystem() 

    val emailServiceRef: ActorRef = system.actorOf(RoundRobinPool(rate).withSupervisorStrategy(supervisorStrategy).props(Props(new EmailService)), "emailWorkerPool") 
    val logServiceRef = system.actorOf(RoundRobinPool(1).props(Props(new NullLogService)), "logWorker") 
    val email = new Email(stringMap.apply("USEREMAIL"), stringMap.apply("REPLYTO"), stringMap.apply("SUBJECT"), stringMap.apply("BODY"), unsubscribeURL) 
    emailServiceRef ! EmailInfo(email, logServiceRef) 
} 

EmailService:

package services 

import akka.actor._ 
import com.amazonaws.services.simpleemail.model.SendRawEmailResult 
import model.Email 
import services.LogService.{Log, LogError} 

object EmailService { 
    case class EmailInfo(email: Email, logServiceRef: ActorRef) 
} 

class EmailService extends Actor { 
    import EmailService._ 

    def receive = { 
    case EmailInfo(email, logServiceRef) =>  
     val emailResult = new SendRawEmailResult 
     emailResult.setMessageId("Fake_Sent_OK") 
     val amazonResult = Try(emailResult) 
     logServiceRef ! Log(email, amazonResult) 
} 
} 
+2

它适用于我:'Email to:[email protected]。 AmazonResultSucess:真。 AmazonResult:电子邮件结果' 我唯一能想到的是在发送消息时是否使用同一个类'NullLogService.Log'。 – Jatin

+1

我不得不同意贾丁。它也适用于我。你在使用ActorSystem吗?是否还有其他代码参与? – siebenschlaefer

+0

我改变了我的NullLogService来打印默认情况。它打印日志(到:电子邮件@ email.com,成功({MessageId:Fake_Sent_OK}))'。案例是“日志”,类型是电子邮件。另外,我将第二个参数的类型更改为“Try(_)”。还是行不通。我不知道该怎么办:C –

回答

1

大概你没有删除LogService.LogLogService.LogError,否则你会得到一个编译错误EmailService。在这种情况下NullLogService不应该定义自己的消息,但使用LogService的那些:

package services 

import akka.actor._ 
import com.amazonaws.services.simpleemail.model.SendRawEmailResult 
import model.Email 

import scala.util.Try 
import LogService.{Log, LogError} 

// no object NullLogService unless you need it for something else 

class NullLogService extends Actor { 
    def receive = { 
    case Log(email, amazonResult) => println("Email to: " + email.to + ". AmazonResultSucess:" + amazonResult.isSuccess.toString + ". AmazonResult: " + amazonResult.get.toString) 
    case LogError(email, amazonException) => println("Error: Email to: " + email.to + ". AmazonException:" + amazonException.get.toString) 
    case [email protected]_ => println("Default case: "+default.toString) 
    } 
} 

然后你就可以使用相同的协议(比如,NullLogServiceSlf4jLogServiceInMemoryLogService)不改变不同的服务之间切换每次进口。

+0

我打算接受我的回答,但要感谢您的建议,我接受您的建议。谢谢。代码现在更清洁 –

0

实测值的误差。我不得不改变我正在导入的记录器类。
我现在使用NullLogService,而不是LogService。

package services 

import akka.actor._ 
import com.amazonaws.services.simpleemail.model.SendRawEmailResult 
import model.Email 
//Change this --> import services.LogService.{Log, LogError} <--- 
//To the class you are using --v 
import services.NullLogService.{Log, LogError} 


object EmailService { 
    case class EmailInfo(email: Email, logServiceRef: ActorRef) 
} 

class EmailService extends Actor { 
    import EmailService._ 

    def receive = { 
    case EmailInfo(email, logServiceRef) =>  
     val emailResult = new SendRawEmailResult 
     emailResult.setMessageId("Fake_Sent_OK") 
     val amazonResult = Try(emailResult) 
     logServiceRef ! Log(email, amazonResult) 
} 
} 

这是奇怪的,因为没有运行时错误(如“NullLogService不导入”)