从 Scala 中的类名获取类型 [英] Get type from class name in Scala

问题描述

我想做如下事情:

val factoryType = typeOf[Class.
          forName("com.menith.amw.worksheets." + params("problem") + "ProblemFactory")]
val factory = parse(params("args")).extract[factoryType]

parse 方法允许我通过给它一个 JSON 字符串来获取 case 类的实例，然后我可以通过将预期类型传递给它来使用提取方法.但是，我在从 Class.forName 获取类型时遇到了一些问题.

The parse method allows me to obtain an instance of a case class by giving it a JSON string and I can then use the extract method by passing it the expected type. However I'm having some issues getting the type from Class.forName.

推荐答案

一个可能有效的很好的解决方案是进行多态反序列化.这允许您向 json 添加一个字段(如类型")并允许 Jackson(假设您重新使用一个很棒的 json 解析器(如 Jackson)来代表您找出正确的类型.

A great solution that might work would be to do polymorphic deserialization. This allows you to add a field (like "type") to your json and allow Jackson (assuming you're using an awesome json parser like Jackson) to figure out the proper type on your behalf.

这篇帖子提供了很好的介绍到多态类型.它涵盖了许多有用的情况，包括您无法修改第 3 方代码的情况(这里您添加了一个 Mixin 来注释类型层次结构).

This post gives a great introduction to polymorphic types. It covers many useful cases including the case where you can't modify 3rd party code (here you add a Mixin to annotate the type hierarchy).

最简单的情况最终看起来像这样(所有这些也适用于 Scala 对象——jackson 甚至有一个很棒的 Scala 模块):

The simplest case ends up looking like this (and all of this works great with Scala objects too -- jackson even has a great scala module):

object Test {
  @JsonTypeInfo(
    use = JsonTypeInfo.Id.NAME,
    include = JsonTypeInfo.As.PROPERTY,
    property = "type"
  )
  @JsonSubTypes(Array(
    new Type(value = classOf[Cat], name = "cat"),
    new Type(value = classOf[Dog], name = "dog")
  ))
  trait Animal

  case class Dog(name: String, breed: String, leash_color: String) extends Animal
  case class Cat(name: String, favorite_toy: String) extends Animal

  def main(args: Array[String]): Unit = {
    val objectMapper = new ObjectMapper with ScalaObjectMapper
    objectMapper.registerModule(DefaultScalaModule)

    val dogStr = """{"type": "dog", "name": "Spike", "breed": "mutt",  "leash_color": "red"}"""
    val catStr = """{"type": "cat", "name": "Fluffy", "favorite_toy": "spider ring"}"""

    val animal1 = objectMapper.readValue[Animal](dogStr)
    val animal2 = objectMapper.readValue[Animal](catStr)

    println(animal1)
    println(animal2)
  }
}

这会生成以下输出:

// Dog(Spike,mutt,red)
// Cat(Fluffy,spider ring)

您也可以避免列出子类型映射，但它要求json类型"字段稍微复杂一些.试验一下；你可能会喜欢.像这样定义动物:

You can also avoid listing the subtype mapping, but it requires that the json "type" field is a bit more complex. Experiment with it; you might like it. Define Animal like this:

@JsonTypeInfo(
  use = JsonTypeInfo.Id.CLASS,
  include = JsonTypeInfo.As.PROPERTY,
  property = "type"
)
trait Animal

它会像这样产生(并消耗)json:

And it produces (and consumes) json like this:

/*
{
    "breed": "mutt",
    "leash_color": "red",
    "name": "Spike",
    "type": "classpath.to.Test$Dog"
}
{
    "favorite_toy": "spider ring",
    "name": "Fluffy",
    "type": "classpath.to.Test$Cat"
}
*/

这篇关于从 Scala 中的类名获取类型的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！