在scala中将JSON对象转换为特定键的值列表（播放）

Question

I have a Json object, which is an output of Json.parse, which is of format

{ "main_data" : [ {"a" : 1, "b" :2} , {"a" : 3, "b" : 4}, {"a" : 5, "b" : 6}]}.

I want to create a list out of this Json object with the values of all the "a" key. So in the above example it will be [1,3,5]. As I am new to functional programming the first thing came to my mind was to write a For loop and traverse through the Json object to get the list.

But I was wondering is there a Functional/Scala way to do the above, using Map or flatMap ?

Answer 1

import play.api.libs.json._

val parsed = Json.parse("""{"main_data":[{"a":1,"b":2},{"a":3,"b":4},{"a":5,"b":6}]}""")
val keys   = parsed \\ "a"
val result = keys.flatMap(_.asOpt[Int]) // List(1, 3, 5)

Answer 2

You can use something like this

val json = Json.parse(yourJsonString)
val aKeys = json \\ "a"
// aKeys: Seq[play.api.libs.json.JsValue] = List(1, 3, 5)
// If you need integers instead of JsValues, just use map
val integersList = aKeys.map(x => x.as[Int])

Answer 3

There are plenty of choices. Let's say you have:

import play.api.libs.json._

val txt =
  """
    |{ "main_data" : [ {"a" : 1, "b" :2} , {"a" : 3, "b" : 4}, {"a" : 5, "b" : 6}]}
  """.stripMargin


val json = Json.parse(txt)

First approach if you are interested only in a :

(json \ "main_data")
  .as[List[Map[String, Int]]]
  .flatten
  .foldLeft(List[Int]()){
    case (acc, ("a", i)) => acc :+ i
    case (acc, _) => acc
  }

The second more general:

(json \ "main_data")
  .as[List[Map[String, Int]]]
  .flatten
  .groupBy(_._1)
  .map {
    case (k, list) => k -> list.map(_._2)
  }
  .get("a")

And the result is:

res0: Option[List[Int]] = Some(List(1, 3, 5))