CODEWITHSUNDEEP
jsonscalaplayframework-json
Vishwas
Vishwas

Reputation: 7067

How to convert list data into Json in scala using play framework?

I am using play framework and Scala.

I have following list-

 List(
    (C: - read,1412750402124,46552070),
    (C: - write,1412750402124,433057), 
(E: - read,1412750402124,5435),  
(E: - write,1412750402124,1728),
(F: - read,1412750402124,5435),  
(F: - write,1412750402124,1728), 
(C: - read,1412750402125,46552071), 
(C: - write,1412750402125,433060),
(E: - read,1412750402125,5445), 
(E: - write,1412750402125,1730),
(F: - read,1412750402125,5450),  
(F: - write,1412750402125,1428)
)

I want following output-

    key:[
{"name":"C:-read",
data:[[1412750402124,46552070],[1412750402125,46552071]...]
},
{
"name":"C:-write",
data:[[1412750402124,433057],[1412750402125,433060]...]
},
{
"name":"E:-read",
data:[[1412750402124,5435],[1412750402125,5445]...]
},
{
"name":"E:-write",
data:[[1412750402124,1728],[1412750402124,1730]...]
},
{
"name":"F:-read",
data:[[1412750402124,5435],[1412750402125,5450]...]
},
{
"name":"F:-write",
data:[[1412750402124,1728],[1412750402124,1428]...]
}
]

How do I get above output using scala??

Upvotes: 0

Views: 622

Answers (1)

roterl
roterl

Reputation: 1883

The input List cannot be compile and some info is missing so I make some assumptions.

val lst = List(
  ("C: - read", 1412750402124L, 46552070),
  ("C: - write", 1412750402124L, 433057),
  ("E: - read", 1412750402124L, 5435),
  ("E: - write", 1412750402124L, 1728),
  ("F: - read", 1412750402124L, 5435),
  ("F: - write", 1412750402124L, 1728),
  ("C: - read", 1412750402125L, 46552071),
  ("C: - write", 1412750402125L, 433060),
  ("E: - read", 1412750402125L, 5445),
  ("E: - write", 1412750402125L, 1730),
  ("F: - read", 1412750402125L, 5450),
  ("F: - write", 1412750402125L, 1428))

Since the results is grouped by the first String (eg "C: - read") the first step is to do the grouping:

scala> val groupedData= lst.groupBy(_._1).map{ case (k,v) => k -> v.map(i => List(i._2, i._3))}.toList
groupedData: List[(String, List[List[Long]])] = List((C: - write,List(List(1412750402124, 433057), List(1412750402125, 433060))), (E: - read,List(List(1412750402124, 5435), List(1412750402125, 5445))), (F: - write,List(List(1412750402124

, 1728), List(1412750402125, 1428))), (F: - read,List(List(1412750402124, 5435), List(1412750402125, 5450))), (E: - write,List(List(1412750402124, 1728), List(1412750402125, 1730))), (C: - read,List(List(1412750402124, 46552070), List( 1412750402125, 46552071))))

Note that the results is List[(String, List[List[Long]])]. This will make the serialization easier.
Play has great Json library and it can easily be done with 'Writes':

implicit val itemWrites = new Writes[(String, List[List[Long]])] {
  def writes(t: (String, List[List[Long]])) = Json.obj(
    "name" -> t._1,
    "data" -> t._2)
}

and now it can be serialize:

scala> Json.toJson(groupedData)
res23: play.api.libs.json.JsValue = [{"name":"C: - write","data":[[1412750402124,433057],[1412750402125,433060]]},{"name":"E: - read","data":[[1412750402124,5435],[1412750402125,5445]]},{"name":"F: - write","data":[[1412750402124,1728]

,[1412750402125,1428]]},{"name":"F: - read","data":[[1412750402124,5435],[1412750402125,5450]]},{"name":"E: - write","data":[[1412750402124,1728],[1412750402125,1730]]},{"name":"C: - read","data":[[1412750402124,46552070],[141275040212 5,46552071]]}]

Upvotes: 1

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