arseny arseny - 1 year ago 76
Scala Question

play json in scala: deserializing json with unknown fields without losing them

consider i have a json as following:

"a": "aa",
"b": "bb",
"c": "cc",
"d": "dd", // unknown in advance
"e": { //unknown in advance
"aa": "aa"

i know for sure that the json will contain a,b,c but i've no idea what other fields this json may contain.

i want to serialize this JSON into a case class containing a,b,c but on the other hand not to lose the other fields (save them in a map so the class will be deserialized to the same json as received).


Answer Source

One option is to capture the "unknown" fields in a Map[String,JsValue], from which you can later extract values if you need them.

case class MyClass(a: String, b: String, c: String, extra: Map[String, JsValue])
implicit val reads: Reads[MyClass] = (
  (__ \ "a").read[String] and
  (__ \ "b").read[String] and
  (__ \ "c").read[String] and[Map[String, JsValue]]
    .map(_.filterKeys(k => !Seq("a", "b", "c").contains(k)))
)(MyClass.apply _)

// Result:
// MyClass(aa,bb,cc,Map(e -> {"aa":"aa"}, d -> "dd"))

Likewise, you can do a Writes or a Format like so:

// And a writes...
implicit val writes: Writes[MyClass] = (
  (__ \ "a").write[String] and
  (__ \ "b").write[String] and
  (__ \ "c").write[String] and
  __.write[Map[String, JsValue]]
)(unlift(MyClass.unapply _))

// Or combine the two...
implicit val format: Format[MyClass] = (
  (__ \ "a").format[String] and
  (__ \ "b").format[String] and
  (__ \ "c").format[String] and
  __.format[Map[String, JsValue]](Reads
    .map[JsValue].map(_.filterKeys(k => !Seq("a", "b", "c").contains(k))))
)(MyClass.apply, unlift(MyClass.unapply))

Note: it looks a bit confusing because you give the format for Map[String,JsValue] an explicit Reads as an argument (, which you then transform (using the .map method) to remove the already-captures values.

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