Scala：如何按名称动态访问类属性？
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时间：2020-10-22 06:55:07 来源:igfitidea点击:
Scala: How to access a class property dynamically by name?
提问by jbrown
How can I look up the value of an object's property dynamically by name in Scala 2.10.x?
如何在 Scala 2.10.x 中按名称动态查找对象属性的值？
E.g. Given the class (it can't be a case class):
例如，给定类（它不能是案例类）：
class Row(val click: Boolean,
val date: String,
val time: String)
I want to do something like:
我想做类似的事情：
val fields = List("click", "date", "time")
val row = new Row(click=true, date="2015-01-01", time="12:00:00")
fields.foreach(f => println(row.getProperty(f))) // how to do this?
回答by Kamil Domański
class Row(val click: Boolean,
val date: String,
val time: String)
val row = new Row(click=true, date="2015-01-01", time="12:00:00")
row.getClass.getDeclaredFields foreach { f =>
f.setAccessible(true)
println(f.getName)
println(f.get(row))
}
回答by Rich Henry
You could also use the bean functionality from java/scala:
您还可以使用 java/scala 中的 bean 功能：
import scala.beans.BeanProperty
import java.beans.Introspector
object BeanEx extends App {
case class Stuff(@BeanProperty val i: Int, @BeanProperty val j: String)
val info = Introspector.getBeanInfo(classOf[Stuff])
val instance = Stuff(10, "Hello")
info.getPropertyDescriptors.map { p =>
println(p.getReadMethod.invoke(instance))
}
}
