JSON Reads combinator in Play scala 2.x doesn't work for Map[String, _]

Question

I have a Play-scala v2.3 application. Looking from this guide about Json Combinators, I'm trying to do this:

object Application extends Controller {

  case class Foo(id: String, docs: List[Map[String, _]])

  implicit val fooReads = (
    (JsPath \ "id").read[String] and
    (JsPath \ "docs").read[List[Map[String, _]]]
  )(Foo.apply _)

  ...
}

But then I got compile error:

No Json deserializer found for type List[Map[String, _]]. Try to implement an implicit Reads or Format for this type.
       (JsPath \ "docs").read[List[Map[String, _]]]
                             ^

This is the example json that need to be read:

{
    "id": "001",
    "docs": [
        {
            "name": "Billy",
            "level": 2,
            "marked": false
        },
        {
            "name": "Fred",
            "level": 5,
            "marked": true
        }
    ]
}

I also have tried this:

  case class Foo(id: String, docs: Map[String, _])

  implicit val fooReads = (
    (JsPath \ "id").read[String] and
    (JsPath \ "docs").read[Map[String, _]]
  )(Foo.apply _)

Same error too.

It seems Play's JSON combinator doesn't work for Map type. Anyone know how to solve this?

It can work for the Map type, but not Map[String, _]. The _ is untyped, so it won't know what to do with it. Why are you trying to use Map[String, _] in the first place? — Michael Zajac
– Michael Zajac, Commented Mar 9, 2015 at 19:51
@m-z: I read from here, it said Json Macros are known to accept Option/Seq/List/Set & Map[String, _], could it be a hint that it should work (at past?). The reason to use Map because I have a service method that accept List[Map[String, _]] parameter, although I can use a list of case class for the docs, but then I have to convert it to List[Map[String, _]]. — null
– null, Commented Mar 9, 2015 at 19:53
I don't think it's meant to literally say Map[String, _], but Map[String, Int], Map[String, String], etc.. for types that make sense. — Michael Zajac
– Michael Zajac, Commented Mar 9, 2015 at 19:56
So it's impossible? The docs json doesn't have uniform type value, that's why I have to use Map[String, _]. — null
– null, Commented Mar 9, 2015 at 19:58
Like that it's not possible without creating your own custom Reads/Writes for it. If they don't have a uniform type, then what do they have? It's still possible to make it work in some way, just not with the macro. — Michael Zajac
– Michael Zajac, Commented Mar 9, 2015 at 20:00

Travis Brown · Accepted Answer · 2015-03-09 21:07:29Z

You'd be a lot better off avoiding Map[String, Any], and Scala and Play make this easy. A much better solution is to define a custom case class for the thing you're trying to represent as a map:

import play.api.libs.json._

case class Doc(name: String, level: Int, marked: Boolean)
case class Foo(id: String, docs: List[Doc])

implicit val docFormat = Json.format[Doc]
implicit val fooFormat = Json.format[Foo]

val json = Json.parse(...)

And then:

scala> json.as[Foo]
res0: Foo = Foo(001,List(Doc(Billy,2,false), Doc(Fred,5,true)))

Or if you want more control:

import play.api.libs.functional.syntax._
import play.api.libs.json._

implicit val fooReads: Reads[Foo] = (
  (__ \ 'id).read[String] and
  (__ \ 'docs).read(
    Reads.list((
      (__ \ 'name).read[String] and
      (__ \ 'level).read[Int] and
      (__ \ 'marked).read[Boolean]
    )(Doc.apply _))
  )
)(Foo.apply _)

If you really need a Map[String, Any], you can always write a conversion method on Doc.

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