如何使用默认值处理选项

coo*_*eze 0 scala

val a = user.type match {
  case Member => doSomething(if(user.location.isDefined) user.location.get.name else "")
}
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我想安全地访问user.location,如果它存在,则为Option [Location],否则只使用空字符串.

case class Location(id: Int, name: String)
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这可能吗?

Jea*_*art 5

只是:

 doSomething(user.location.map(_.name).getOrElse(""))
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