如何访问访问案例类字段值来自字段的字符串名称

Question

如何从表示该字段的给定String值中提取案例类的字段值.

例如:

case class Person(name: String, age: Int)
val a = Person("test",10)

现在这里给出一个字符串name 或age我想从变量中提取值a.我该怎么做呢？我知道这可以用反射完成,但我不确定如何？

Answer 1

您正在寻找的是使用无形镜片.这也将在编译时而不是运行时将字段实际存在于案例类的约束:

import shapeless._

case class Person(name: String, age: Int)

val nameLens = lens[Person] >> 'name
val p = Person("myName", 25)

nameLens.get(p)

产量:

res0: String = myName

如果您尝试提取非现有字段,则会出现编译时错误,这是一个更强大的保证:

import shapeless._

case class Person(name: String, age: Int)

val nonExistingLens = lens[Person] >> 'bla
val p = Person("myName", 25)

nonExistingLens.get(p)

编译器大叫:

Error:(5, 44) could not find implicit value for parameter mkLens: shapeless.MkFieldLens[Person,Symbol with shapeless.tag.Tagged[String("bla")]]
val nonExistingLens = lens[Person] >> 'bla

Answer 2

不知道你到底在想什么，但一个match语句会做，它对于Person案例类的更改不是很通用或可扩展，但它确实满足你不使用反射的基本要求：

scala> val a = Person("test",10)
a: Person = Person(test,10)

scala> def extract(p: Person, fieldName: String) = {
     |   fieldName match {
     |     case "name" => p.name
     |     case "age" => p.age
     |   }
     | }
extract: (p: Person, fieldName: String)Any

scala> extract(a, "name")
res1: Any = test

scala> extract(a, "age")
res2: Any = 10

scala> extract(a, "name####")
scala.MatchError: name#### (of class java.lang.String)
  at .extract(<console>:14)
  ... 32 elided

根据评论更新：

scala> case class Person(name: String, age: Int)
defined class Person

scala> val a = Person("test",10)
a: Person = Person(test,10)


scala> def extract(p: Person, fieldName: String) = {
     |   fieldName match {
     |     case "name" => Some(p.name)
     |     case "age" => Some(p.age)
     |     case _ => None
     |   }
     | }
extract: (p: Person, fieldName: String)Option[Any]

scala> extract(a, "name")
res4: Option[Any] = Some(test)

scala> extract(a, "age")
res5: Option[Any] = Some(10)

scala> extract(a, "name####")
res6: Option[Any] = None

scala>