Question

我有一个灵感来自spala scala demo here的API类，我写的是在喷射路径中将Person渲染为JSON对象。

trait UsersApi {

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

  object MyJsonProtocol extends DefaultJsonProtocol {
    implicit val PersonFormat = jsonFormat3(Person)
  }

  import MyJsonProtocol._
  import spray.httpx.SprayJsonSupport._
  import spray.util._

  val bob = Person("Bob", "Parr", 32)

  val usersApiRouting: Route = {
    path("users") {
      get {
        complete {
          marshal(bob)
        }
      }
    }
  }
}

问题是编组（bob）像这样返回JSON：

{
  "name": "Bob",
  "firstName": "Parr",
  "age": 32
}

想象一下，我需要像这样渲染没有“年龄”的JSON：

{
  "name": "Bob",
  "firstName": "Parr"
}

如何实现这一目标？我有一个想法是Scala有办法使一个对象成为另一个对象属性的子集吗？或者也许喷-json有一些特定的支持，不编组不应该添加到服务器响应的属性？

Answer 1

根据spray-json docs，你应该像这样提供自定义的jsonFormat：

case class Person(name: String, firstName: String, age: Option[Int])

  object MyJsonProtocol extends DefaultJsonProtocol {
    implicit object PersonFormat extends RootJsonFormat[Person] {
    def write(p: Person) =JsObject(
      "name" -> JsString(p.name),
      "firstName" -> JsString(p.firstName)
    )

    def read(value: JsValue) = {
      value.asJsObject.getFields("name", "firstName", "age") match {
        case Seq(JsString(name), JsString(firstName), JsNumber(age)) =>
          new Person(name, firstName, Some(age.toInt))
        case Seq(JsString(name), JsString(firstName)) =>
          new Person(name, firstName, None)
        case _ => throw new DeserializationException("Person expected: " + value.asJsObject.getFields("name", "firstName", "age").toString)
      }
    }
  }
  }

  import MyJsonProtocol._
  import spray.httpx.SprayJsonSupport._
  import spray.util._

  val bob = Person("Bob", "Parr", Some(32))

  val bobJson = bob.toJson.toString  //bobJson: String = {"name":"Bob","firstName":"Parr"}

  val bobObj = bobJson.parseJson.convertTo[Person]  //bobObj: Person = Person(Bob,Parr,None)

使用spray-json中的对象属性子集返回json响应

1 个答案: