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

问题

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": ""
}

回答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)))

回答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)

回答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

来源：https://stackoverflow.com/questions/24512942/play-how-to-remove-the-fields-without-value-from-json-and-create-a-new-json-wit

标签

json

scala

playframework

play-json