Play Framework - add a field to JSON object

21,411

Solution 1

JsObject has a + method that allows you to add fields to an object, but unfortunately your jsonObject is statically typed as a JsValue, not a JsObject. You can get around this in a couple of ways. The first is to use as:

 scala> jsonObject.as[JsObject] + ("c" -> Json.toJson(3))
 res0: play.api.libs.json.JsObject = {"a":1,"b":2,"c":3}

With as you're essentially downcasting—you're telling the compiler, "you only know that this is a JsValue, but believe me, it's also a JsObject". This is safe in this case, but it's not a good idea. A more principled approach is to use the OWrites directly:

scala> val jsonObject = classAWrites.writes(classAObject)
jsonObject: play.api.libs.json.JsObject = {"a":1,"b":2}

scala> jsonObject + ("c" -> Json.toJson(3))
res1: play.api.libs.json.JsObject = {"a":1,"b":2,"c":3}

Maybe someday the Json object will have a toJsonObject method that will require a OWrites instance and this overly explicit approach won't be necessary.

Solution 2

I found a solution myself. In fact the JsValue, which is the return type of Json.toJson has no such method, but the JsObject (http://www.playframework.com/documentation/2.2.x/api/scala/index.html#play.api.libs.json.JsObject) does, so the solution is:

val jsonObject = Json.toJson(classAObject).as[JsObject]
jsonObject + ("c", JsNumber(3)) 

I hope someone will find this useful :)

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Paweł Kozikowski
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Paweł Kozikowski

Updated on May 23, 2020

Comments

  • Paweł Kozikowski
    Paweł Kozikowski almost 4 years

    I have a problem with adding a field to Json object in Play Framework using Scala:

    I have a case class containing data. For example:

    case class ClassA(a:Int,b:Int)
    

    and I am able to create a Json object using Json Writes:

    val classAObject = ClassA(1,2)
    implicit val classAWrites= Json.writes[ClassA]
    val jsonObject = Json.toJson(classAObject)
    

    and the Json would look like:

    { a:1, b:2 }
    

    Let's suppose I would like to add an additional 'c' field to the Json object. Result:

    { a:1, b:2, c:3 }
    

    How do I do that without creating a new case class or creating my Json object myself using Json.obj? I am looking for something like:

    jsonObject.merge({c:3}) 
    

    Any help appreciated!

  • Tvaroh
    Tvaroh over 9 years
    Per signature, Writes.writes returns JsValue, so I don't understand how you can get rid of upcasting (not in the REPL).
  • elmalto
    elmalto over 9 years
    Tvaroh is right, it return JsValue as far as I can see
  • Travis Brown
    Travis Brown over 9 years
    @elmalto Try it out—classAWrites will be statically typed as OWrites[ClassA] (in both Play 2.2 and 2.3 and on 2.10 and 2.11). This is due to "underspecified but intended" behavior of Scala's macros (see my question here for details).
  • elmalto
    elmalto over 9 years
    Skipping the definition for the class and writer for brevity, but this is what the repl gives me: scala> SessionCreatorWriter.writes(s) res2: play.api.libs.json.JsValue
  • Travis Brown
    Travis Brown over 9 years
    @elmalto If you define a Writes instance you'll get JsValue, but if you define an OWrites instance—either manually or using the Json.writes macro, which is what the OP is doing—you'll get a JsObject.
  • elmalto
    elmalto over 9 years
    @TravisBrown maybe I should've posted it then: implicit val SessionCreatorWriter = Json.writes[SessionCreator] val s = SessionCreator("User1",None,None,None,1,None,None) SessionCreatorWriter.writes(s) and I get a JsValue
  • Wrench
    Wrench almost 9 years
    There's actually no need for the paranthesis. This works: jsobj + "key" -> entity.key
  • Travis Brown
    Travis Brown almost 9 years
    @Wrench That does something completely different. + and - have the same precedence, so the parentheses are definitely necessary.
  • Basil
    Basil over 3 years
    Why is Json.toJson(3) needed in this scenario. Even though it's not needed when doing Json.obj("a" -> 3) ??