在可能出现异常的地方应用未来

Question

这段代码运行良好，但我想通过Future 管理线程。 sendSMS 方法执行通常需要 3 到 5 秒，我想申请未来并申请在一个地方但想知道它是否足够？

val c = for {
            t <- Future { doSendSms("+9178787878787","i scare with threads") }
          } yield t
c.map { res =>
        res match {
          case e: Error => {
            Ok(write(Map("result" -> "error")))
          }
          case Success() => {
            Ok(write(Map("result" -> "success")))
          }

def doSendSms(recipient: String, body: String): SentSmsResult = {
    try {
      sendSMS(recipient, body)
      Success()
    } catch {
      case twilioEx: TwilioRestException =>
          return Error(twilioEx.toString)
      case e: Exception =>
          return Error(e.toString)
    }
  }


def sendSMS(smsTo: String, body: String) = {
    val params = Map("To" -> smsTo, "From" -> twilioNumber, "Body" -> body)
    val messageFactory = client.getAccount.getSmsFactory
    messageFactory.create(params)
  }// sending sms from twilio, this method takes 3 to 5 seconds to execute

如果不是如何在这段代码中管理 Future

Answer 1

我会使用recover:

val c = for {
  t <- doSendSms("+9178787878787","i scare with threads")
} yield t

def doSendSms(recipient: String, body: String): Future[SentSmsResult] = 
  Future {
    sendSMS(recipient, body)
  }
  .recover {
    case twilioEx: TwilioRestException => Error(twilioEx.toString)
    case e: Exception => Error(e.toString)
  }
}

recover 将捕获未来执行中引发的异常，从而允许您返回包装在 Future 中的新结果，如文档所述：

如果成功完成，recover 组合器会创建一个新的未来，它与原始未来保持相同的结果。如果没有，则将部分函数参数应用于原始未来失败的 Throwable。