CODEWITHSUNDEEP
jsonscalaplayframeworkplayframework-2.0jackson
Caballero
Caballero

Reputation: 12091

Scala/Play: JSON serialization issue

I have a simple custom data structure which I use to map the results from the database:

case class Filter(id: Int, table: String, name: String, Type: String, structure: String)

The resulting object type is List[Filter] and if converted to JSON, it should look something like this:

[
    {
        "id":           1,
        "table":        "table1",
        "name":         "name1",
        "Type":         "type1",
        "structure":    "structure1"
    },
    {
        "id":           2,
        "table":        "table2",
        "name":         "name2",
        "Type":         "type2",
        "structure":    "structure2"
    }
]

Now when I try to serialize my object into JSON

val result: String = Json.toJson(filters)

I am getting something like

No Json deserializer found for type List[Filter]. Try to implement an implicit Writes or Format for this type.

How do I solve this seemingly simple problem without writing some ridiculous amount of boilerplate?

My stack is Play 2.2.1, Scala 2.10.3, Java 8 64bit

Upvotes: 3

Views: 1644

Answers (1)

thesamet
thesamet

Reputation: 6582

Short answer:

Just add:

implicit val filterWrites = Json.writes[Filter]

Longer answer:

If you look at the definition of Json.toJson, you will see that its complete signature is:

def toJson[T](o: T)(implicit tjs: Writes[T]): JsValue = tjs.writes(o)

Writes[T] knows how to take a T and transform it to a JsValue. You will need to have an implicit Writes[Filter] around that knows how to serialize your Filter instance. The good news is that Play's JSON library comes with a macro that can instantiate those Writes[_] for you, so you don't have to write boring code that transforms your case class's fields into JSON values. To invoke this macro and have its value picked up by implicit search add the line above to your scope.

Upvotes: 2

Related Questions