播放Scala JSON：合并属性

Question

I have the following case class:

case class User(name: String) .

I am trying to implement a JSON Reads converter for it, so I can do the following:

val user = userJson.validate[User]

… but the incoming JSON has slightly different structure:

{ "firstName": "Bob", "lastName": "Dylan" } .

How can I implement my JSON Reads converter to combine the JSON fields firstName and lastName into a name property on my class?

Answer 1

This should do the trick:

  implicit val userReads: Reads[User] = 
      for {
        first <- (__ \ "firstName").read[String]
        last <- (__ \ "lastName").read[String]
      } yield User(s"$first $last")

EDIT Without using a for comprehension

implicit val userReads = 
  { 
    (__ \ "firstName").read[String] and 
    (__ \ "lastName"
  }.read[String] ).tupled.map(t => User(s"${t._1} ${t._2}"))

Bringing userReads in scope where you want to use it will let you parse the JSON you provided.

Reads is essentially a function from JsValue to JsResult , meaning userReads represents a function from JsValue -> JsResult . Within the function, it first inspects the provided JSON & tries to read out a property named "firstName" from the current JSON path ( __ is shorthand for this). \\ indicates that the field its looking for is one level beneath the root, and read[String] means the value associated with the "firstName" key should be read as a string. Same follows for "lastName".

Edit In the version without the for comprehension, it first creates an intermediary object FunctionalBuilder[Reads]#CanBuild[String, String] , which is a complicated way of saying it reads two distinct strings from the Json. Next it converts that complex object into a Reads[(String, String)] by way of tupled . Finally it maps the pair of strings into a User .

Were you to try validating some JSON without "firstName" & "lastName", this will fail with a validation error for a missing path.