是否有任何 Scala 功能允许您调用名称存储在字符串中的方法？
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Is there any Scala feature that allows you to call a method whose name is stored in a string?
提问by Geo
Assuming you have a string containing the name of a method, an object that supports that method and some arguments, is there some language feature that allows you to call that dynamically?
假设您有一个包含方法名称的字符串、一个支持该方法的对象和一些参数，是否有某种语言功能允许您动态调用它？
Kind of like Ruby's sendparameter.
有点像 Ruby 的send参数。
回答by Rex Kerr
You can do this with reflection in Java:
您可以使用 Java 中的反射来做到这一点：
class A {
def cat(s1: String, s2: String) = s1 + " " + s2
}
val a = new A
val hi = "Hello"
val all = "World"
val method = a.getClass.getMethod("cat",hi.getClass,all.getClass)
method.invoke(a,hi,all)
And if you want it to be easy in Scala you can make a class that does this for you, plus an implicit for conversion:
如果你希望在 Scala 中它很容易，你可以创建一个为你做这件事的类，加上一个隐式的转换：
case class Caller[T>:Null<:AnyRef](klass:T) {
def call(methodName:String,args:AnyRef*):AnyRef = {
def argtypes = args.map(_.getClass)
def method = klass.getClass.getMethod(methodName, argtypes: _*)
method.invoke(klass,args: _*)
}
}
implicit def anyref2callable[T>:Null<:AnyRef](klass:T):Caller[T] = new Caller(klass)
a call ("cat","Hi","there")
Doing this sort of thing converts compile-time errors into runtime errors, however (i.e. it essentially circumvents the type system), so use with caution.
但是，执行此类操作会将编译时错误转换为运行时错误（即它本质上绕过了类型系统），因此请谨慎使用。
(Edit: and see the use of NameTransformer in the link above--adding that will help if you try to use operators.)
（编辑：并在上面的链接中查看 NameTransformer 的使用 - 如果您尝试使用运算符，添加这将有所帮助。）
回答by Jim Barrows
Yes. It's called reflection. Here's a link to one way, using some experimental stuffHowever you should remember that Scala is not a dynamic language, and may not be able to easily do some things that scripting languages can do. You're probably better doing a match on the string, and then calling the right method.
是的。这叫做反射。 这是一种方法的链接，使用一些实验性的东西但是您应该记住 Scala 不是一种动态语言，并且可能无法轻松完成脚本语言可以做的一些事情。您可能最好对字符串进行匹配，然后调用正确的方法。
回答by Vasily802
Yes you can!You would need .invoke()method of the method object. Simple example below:
是的你可以！您将需要.invoke()方法对象的方法。下面的简单例子：
import scala.util.Try
case class MyCaseClass(i: String) {
def sayHi = {
println(i)
}
}
val hiObj = MyCaseClass("hi")
val mtdName = "sayHi"
// Method itself as an object
val mtd = hiObj.getClass.getMethod(mtdName)
Try {mtd.invoke(hiObj)}.recover { case _ => ()}
see code here: https://scastie.scala-lang.org/vasily802/WRsRpgSoSayhHBeAvogieg/9
在这里查看代码：https: //scastie.scala-lang.org/vasily802/WRsRpgSoSayhHBeAvogieg/9
回答by Ravi
scala> val commandExecutor = Map("cleanup" -> {()=> println("cleanup successfully")} )
commandExecutor: scala.collection.immutable.Map[String,() => Unit] = Map(cleanup -> <function0>)
scala> val command="cleanup"
command: String = cleanup
scala> commandExecutor(command).apply
cleanup successfully
