如何轻松获得 Scala 案例类的名称？
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/2656364/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
How can I easily get a Scala case class's name?
提问by pr1001
Given:
鉴于：
case class FirstCC {
def name: String = ... // something that will give "FirstCC"
}
case class SecondCC extends FirstCC
val one = FirstCC()
val two = SecondCC()
How can I get "FirstCC"from one.nameand "SecondCC"from two.name?
我怎样才能"FirstCC"从one.name和"SecondCC"从two.name？
回答by Esko Luontola
def name = this.getClass.getName
Or if you want only the name without the package:
或者，如果您只想要名称而不需要包：
def name = this.getClass.getSimpleName
See the documentation of java.lang.Classfor more information.
有关更多信息，请参阅java.lang.Class的文档。
回答by Patrick
You can use the property productPrefixof the case class:
您可以使用productPrefix案例类的属性：
case class FirstCC {
def name = productPrefix
}
case class SecondCC extends FirstCC
val one = FirstCC()
val two = SecondCC()
one.name
two.name
N.B.
If you pass to scala 2.8 extending a case class have been deprecated, and you have to not forget the left and right parent ()
NB如果你传递给scala 2.8扩展案例类已被弃用，你必须不要忘记左右父级 ()
回答by Daniel C. Sobral
class Example {
private def className[A](a: A)(implicit m: Manifest[A]) = m.toString
override def toString = className(this)
}
回答by Rex Kerr
def name = this.getClass.getName
回答by eje
Here is a Scala function that generates a human-readable string from any type, recursing on type parameters:
这是一个 Scala 函数，它从任何类型生成一个人类可读的字符串，递归类型参数：
https://gist.github.com/erikerlandson/78d8c33419055b98d701
https://gist.github.com/erikerlandson/78d8c33419055b98d701
import scala.reflect.runtime.universe._
object TypeString {
// return a human-readable type string for type argument 'T'
// typeString[Int] returns "Int"
def typeString[T :TypeTag]: String = {
def work(t: Type): String = {
t match { case TypeRef(pre, sym, args) =>
val ss = sym.toString.stripPrefix("trait ").stripPrefix("class ").stripPrefix("type ")
val as = args.map(work)
if (ss.startsWith("Function")) {
val arity = args.length - 1
"(" + (as.take(arity).mkString(",")) + ")" + "=>" + as.drop(arity).head
} else {
if (args.length <= 0) ss else (ss + "[" + as.mkString(",") + "]")
}
}
}
work(typeOf[T])
}
// get the type string of an argument:
// typeString(2) returns "Int"
def typeString[T :TypeTag](x: T): String = typeString[T]
}
