CODEWITHSUNDEEP
scalaplayframeworksalat
user2066049
user2066049

Reputation: 1371

How to convert JSON into Scala class (NOT Case class), and then populate set of case classes from that big class

I am building an application using Scala 2.10, Salat and Play frmework 2.1-RC2 (will upgrade to 2.1 release soon) and MongoDB.

This is a faceless application where JSON web services are exposed for consumers. Up until now JSON was converted into Model object directly using Play's Json API and implicit converters. I have to refactor some case classes to avoid 22 tuples limit and now instead of flat case class I'm now refactoring to have an embedded case(and embedded MongoDB collection).

Web service interface should remain same where client should still be passing in JSON data as they were before in a flat structure but application needs to map them into proper case class(es) structure. What's the best way to handle this kind of situation. I fear of writing a lot of conversion code <-> Flat JSON <-> complex case class structure <-> from complex case classes to flat JSON output again.

How would you approach such a requirement? I assume case class 22 tuple limit may have had been faced by many others to handle this kind of requirements? How would you approach this

Upvotes: 2

Views: 1157

Answers (1)

EECOLOR
EECOLOR

Reputation: 11244

The Play 2.1 json library relies heavily on combinators (path1 and path2). These combinators all have the same 22 restriction. That gives you two options:

  1. Don't use combinators and construct your objects the hard way: path(json) will give you the value at that point in the path. Searching for 'Accessing value of JsPath' at ScalaJsonCombinators will give more examples.
  2. First transform the json into a structure that does not have more than 22 values in a single object and then use the normal combinators. More information about transforming can be found here: ScalaJsonTransformers

Upvotes: 1

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