Scala unpickle with missing field

Question

How can I unpickle an object with a field that might miss (and use an null) in that case?

scala> case class Person(name:String=null,age:Int)
defined class Person
scala> import scala.pickling.Defaults._, scala.pickling.json._
scala> val p2 = JSONPickle("""{"age":2}""").unpickle[Person]
scala.pickling.PicklingException: No field 'name' when unpickling, tag Person, fields were Map(age -> 2.0)

I need this in order to unpickle a string came from browser that is also missing the type field, so I will not be able to use an Option/Some/None type.

@jsuereth recommands https://github.com/sbt/serialization . I will take a look. — raisercostin, Jul 30 '15 at 06:42

score 1 · Answer 1 · answered Jul 25 '15 at 12:14

1

Just declare it having type Option[String], like this:

case class Person(name: Option[String], age: Int)

Option[T] is the idiomatic Scala way of handling values which otherwise (e.g. in Java) could be null

answered Jul 25 '15 at 12:14

Denis Mikhaylov

2,035
21
24

Unfortunately this was not an option since the web browser will need to create the json for Person and knowing the pickling internal format might be a little to much to ask. I was hoping that a formatter to plain simple json exists. – raisercostin Jul 29 '15 at 05:11
If you browser just don't send `name` field it will be treated as `None`. – Denis Mikhaylov Jul 30 '15 at 06:03
I tested with Option[String] and it seems that it still fails if the name is missing. Besides what string should come from browser to have a Some("name1") as name? The browser should send something like { "$type": "Person", "name": { "$type": "scala.Some[java.lang.String]", "x": "name1" }, "age": 1 } which is not something too elegant/robust. – raisercostin Jul 30 '15 at 06:40
Absolutely not! You may just send `{"$type": "Person", "name": "name1, "age": 1}` or `{"$type": "Person", "age": 1}` if name is not present. – Denis Mikhaylov Jul 30 '15 at 07:04
Can you share the imports that you are using to parse that string? I used `import scala.pickling.Defaults._, scala.pickling.json._` and it fails with `scala> val p2 = JSONPickle("""{"$type": "Person", "name": "name1", "age": 1}""").unpickle[Person] scala.MatchError: name1 (of class java.lang.String)` – raisercostin Jul 30 '15 at 12:25

score 1 · Answer 2 · edited May 23 '17 at 12:22

To my understanding, scala-pickling can not handle missing fields.

If we look at the output of how scala-pickle pickles Option:

Some(9).pickle
// JSONPickle({
//   "$type": "scala.Some[scala.Int]",
//   "x": 9
// })

case class TestA(x: Option[Int])
TestA(None).pickle
// JSONPickle({
//   "$type": "TestA",
//   "x": {
//     "$type": "scala.None.type"
//   }
// })

You could use a json library instead of a serialization library.

Given your json and case class with an optional name:

case class Person(name:Option[String], age:Int)
val json =  """{"age":2}"""

With json4s:

import org.json4s._
import org.json4s.JsonDSL._
import org.json4s.native.JsonMethods._

parse(json).extract[Person]
// Person = Person(None,2)

With argonaut :

import argonaut._, Argonaut._

implicit def PersonCodecJson = 
  casecodec2(Person.apply, Person.unapply)("name", "age")

json.decodeOption[Person]
// Option[Person] = Some(Person(None,2))

Scala unpickle with missing field

2 Answers2