json - 将 circe 中 json 对象的所有键从 `underscore` 转换为 `camel case`

Question

起源

{
  "first_name" : "foo",
  "last_name" : "bar",
  "parent" : {
    "first_name" : "baz",
    "last_name" : "bazz",
  }
}

预期的

 {
      "firstName" : "foo",
      "lastName" : "bar",
      "parent" : {
        "firstName" : "baz",
        "lastName" : "bazz",
      }
    }

如何转换 json 对象的所有键？

Answer 1

这就是我写这个的方式。它没有我想要的那么简洁，但并不可怕:

import cats.free.Trampoline
import cats.std.list._
import cats.syntax.traverse._
import io.circe.{ Json, JsonObject }

/**
 * Helper method that transforms a single layer.
 */
def transformObjectKeys(obj: JsonObject, f: String => String): JsonObject =
  JsonObject.fromIterable(
    obj.toList.map {
      case (k, v) => f(k) -> v
    }
  )

def transformKeys(json: Json, f: String => String): Trampoline[Json] =
  json.arrayOrObject(
    Trampoline.done(json),
    _.traverse(j => Trampoline.suspend(transformKeys(j, f))).map(Json.fromValues),
    transformObjectKeys(_, f).traverse(obj => Trampoline.suspend(transformKeys(obj, f))).map(Json.fromJsonObject)
  )

进而:

import io.circe.literal._

val doc = json"""
{
  "first_name" : "foo",
  "last_name" : "bar",
  "parent" : {
    "first_name" : "baz",
    "last_name" : "bazz"
  }
}
"""

def sc2cc(in: String) = "_([a-z\\d])".r.replaceAllIn(in, _.group(1).toUpperCase)

最后:

scala> import cats.std.function._
import cats.std.function._

scala> transformKeys(doc, sc2cc).run
res0: io.circe.Json =
{
  "firstName" : "foo",
  "lastName" : "bar",
  "parent" : {
    "firstName" : "baz",
    "lastName" : "bazz"
  }
}

我们可能应该有一些递归应用 Json => F[Json] 的方法。这样转换更方便。