Play: How to remove the fields without value from JSON and create a new JSON with them

Question

Given the following JSON:

{
  "field1": "value1",
  "field2": "",
  "field3": "value3",
  "field4": ""
}

How do I get two distinct JSONs, one containing the fields with value and another one containing the fields without value? Here below is how the final result should look like:

{
  "field1": "value1",
  "field3": "value3"
}

{
  "field2": "",
  "field4": ""
}

Spinal

and just looping through and building two new objects isn't sufficient? – Guncotton

Answer 1

You have access to the JSON object's fields as a sequence of (String, JsValue) pairs and you can filter through them. You can filter out the ones with and without value and use the filtered sequences to construct new JsObject objects.

import play.api.libs.json._

val ls =
  ("field1", JsString("value1")) ::
  ("field2", JsString("")) ::
  ("field3", JsString("value3")) ::
  ("field4", JsString("")) ::
  Nil

val js0 = new JsObject(ls)

def withoutValue(v: JsValue) = v match {
  case JsNull => true
  case JsString("") => true
  case _ => false
}

val js1 = JsObject(js0.fields.filterNot(t => withoutValue(t._2)))
val js2 = JsObject(js0.fields.filter(t => withoutValue(t._2)))

Beason

Answer 2

You can improve @nietaki solution by using partition function.

import play.api.libs.json._

val ls =
  ("field1", JsString("value1")) ::
    ("field2", JsString("")) ::
    ("field3", JsString("value3")) ::
    ("field4", JsString("")) ::
    Nil

val js0 = new JsObject(ls)

def withoutValue(v: JsValue) = v match {
  case JsNull => true
  case JsString("") => true
  case _ => false
}

val (js1, js2) = js0.fields.partition(t => withoutValue(t._2))
JsObject(js1)
JsObject(js2)

Peeling

Answer 3

You can also do it in a one liner:

val (js1, js2) = j.fields.partition(_._2 != JsString(""))
println(JsObject(js1))
println(JsObject(js2))

Code run at Scastie

Lohman