how to convert hashMap into json object in scala?

Question

I want to convert hashmap into json object, my hashmap structure is look like this:

def res=Action{ implicit request=>
  var response=new HashMap[String,Map[String,String]]
  response=//etc .......
  .
  .
  .
  Ok(write(response))
}

bt its not working .

Take a look at the json utilities built into play framework, then apply the knowledge — om-nom-nom
– om-nom-nom, Commented Apr 7, 2014 at 11:22
it depends upon your requirement... can you show me which type format u want to get ? — Ashish Ratan
– Ashish Ratan, Commented Apr 8, 2014 at 12:06

serejja · Accepted Answer · 2014-04-07 11:24:01Z

4

Try this:

Ok(Json.toJson(response.toMap))

This would convert your HashMap to a Map which can be written as json without additional code.

answered Apr 7, 2014 at 11:24

serejja

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8 Comments

Rohit Sharma Over a year ago

i tried it but it giving me can not write an instanse of com.fasterxml.jackson.databind.jsonNode to http response try to define a writable[com.fasterxml.jackson.databind.jsonNode]

serejja Over a year ago

It seems you imported the wrong package. Make sure you imported play.api.libs.json.Json

Rohit Sharma Over a year ago

its not working properly its giving me {"keyValue":"LoadFactor:0.75"....not giving key's Value

Rohit Sharma Over a year ago

import net.liftweb.json.Serialization.write ,import net.liftweb.json.DefaultFormats, import play.api.libs.json.Json

serejja Over a year ago

Why would you use java.util.Map? scala.collection.immutable.Map is the only Map that can be serialized without implementing writes or formats

|

Andreas Neumann · Accepted Answer · 2014-04-07 14:07:23Z

3

An alternative solution would be to use JSON4s. [https://github.com/json4s/json4s] As additional gain it gives you a nice DSL, the abilitity to use Jackson or not and a great way to deserialize JSON.

scala> import org.json4s.jackson.Serialization
import org.json4s.jackson.Serialization

scala> implicit val formats = Serialization.formats(NoTypeHints)
formats: org.json4s.Formats{val dateFormat: org.json4s.DateFormat; val typeHints:org.json4s.TypeHints} = org.json4s.Serialization$$anon$1@f40c08d

scala> Serialization.write(Map("test" -> "a", "test 2" -> 2))
res1: String = {"test":"a","test 2":2}

answered Apr 7, 2014 at 14:07

Andreas Neumann

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Comments

S.Karthik · Accepted Answer · 2014-04-07 12:05:32Z

1

Try like this

val data = response.map(value=> value._1 -> Json.toJson(value._2))

Ok(json.toJson(data.toMap))

edited Apr 7, 2014 at 12:05

answered Apr 7, 2014 at 12:03

S.Karthik

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2 Comments

Rohit Sharma Over a year ago

its giving compile time error ............No json deserializer found for type java.util.map[String,String] try implement implicit write or format for this type.

S.Karthik Over a year ago

then try convert map to list like response.map(value=> value._1 -> Json.toJson(value._2.toList))

Ashish Ratan · Accepted Answer · 2014-04-08 12:20:50Z

0

You can try like this....

var ls: ListBuffer[(String, Map[String, String])] = ListBuffer()
val res = list1.toList.iterator

while (res.hasNext) {

  ls += (("id", getMyMap().toMap))
}
println(ls);
ls.toList
ok(write(ls.toMap))


def getMyMap(): scala.collection.mutable.Map[String, String] = {

var m=scala.collection.mutable.Map("Address" -> "strt1", "Mobile" -> 98974)
 m
}

Output:

{"0":{"Address":"strt1","Mobile":"98974"}}

answered Apr 8, 2014 at 12:20

Ashish Ratan

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