小编Kla*_*aus的帖子

我正在尝试在Miles Sabin的博客文章中定义的Scala的联合类型:

http://www.chuusai.com/2011/06/09/scala-union-types-curry-howard/

并在中讨论过

如何定义"类型析取"(联合类型)？

对于那里定义的简单情况,它们工作正常,但我要做的是使用它们在Play Framework中创建一个只接受某些值(String,Boolean,Int,Undefined)的通用JSON解析器,然后成功传递它们.

这是我的代码:

type UpdateType = Option[String] |?| Option[Int] |?| Option[Boolean] |?| Option[List[Int]]

def withValue[T : (UpdateType)#?](request: Request[JsValue])(block: (String, T) => Future[SimpleResult]) = {
  val field = request.body \ ("field")
  val value = request.body \ ("value")
  (field, value) match {
    case (x: JsString, y: JsString) => block(x.value.toString, Some(y.value.toString))
    case (x: JsString, y: JsNumber) => block(x.value.toString, Some(y.value.intValue))
    case (x: JsString, y: JsBoolean) => block(x.value.toString, Some(y.value.booleanValue))
    case (x: JsString, y: JsUndefined) => block(x.value.toString, None)
    case _ => …