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

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": ""
}
like image 499
j3d Avatar asked Jul 01 '14 14:07

j3d


2 Answers

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)))
like image 50
nietaki Avatar answered Sep 22 '22 10:09

nietaki


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)
like image 43
Przemek Avatar answered Sep 21 '22 10:09

Przemek