Question

我正在使用Scala＆amp;玩2.5。我坚持这个错误：

Game.scala:99: overloaded method value apply with alternatives:
[error]   (block: => play.api.mvc.Result)play.api.mvc.Action[play.api.mvc.AnyContent] <and>
[error]   (block: play.api.mvc.Request[play.api.mvc.AnyContent] => play.api.mvc.Result)play.api.mvc.Action[play.api.mvc.AnyContent] <and>
[error]   [A](bodyParser: play.api.mvc.BodyParser[A])(block: play.api.mvc.Request[A] => play.api.mvc.Result)play.api.mvc.Action[A]
[error]  cannot be applied to (Object)
[error]     def start(id: String, apiKey: Option[String]) = Action {

这是功能：

def start(id: String, apiKey: Option[String]) = Action {
  apiKey match {
    case Some(API_KEY) => {
      Server.actor ! Server.Start(id)
      Ok("Started")
    }
    case _ => Future.successful(Unauthorized)
  }
}

Answer 1

问题是，match语句的结果被推断为Object，因为从一个案例陈述中您返回Result，而另一个案例陈述则是Future[Result]返回Object，因此唯一常见的超类型是case _ => Future.successful(Unauthorized)。要解决此问题，请将case _ => Unauthorized更改为from pyspark.sql import HiveContext HiveContext(sc).sql('from `dbname.tableName` select `*`')。

Scala / Play 2.5：重载方法适用于替代方案

1 个答案: