3

我有一些历史数据,我想将其转换为 json。所以这些是列表列表。它们的类型是 List[List[Position]] 其中 Position 是一个简单的案例类。我写了一个格式化程序来帮助 Json.toJson 应对。我期待一个带有两个内部数组的外部数组的输出,每个内部数组包含 3 个对象。我得到的是这个。请注意额外的数组包装。

[[[[{"amount":1.0,"minAmount":2.0,"price":3.0,"volume":4.0},
{"amount":5.0,"minAmount":6.0,"price":7.0,"volume":8.0},
{"amount":9.0,"minAmount":10.0,"price":11.0,"volume":12.0}]],
[[{"amount":0.1,"minAmount":0.2,"price":0.3,"volume":0.4},
{"amount":5.0,"minAmount":6.0,"price":7.0,"volume":8.0},
{"amount":9.0,"minAmount":10.0,"price":11.0,"volume":12.0}]]]]

我不知道包装数组来自哪里。有人可以帮我吗?这是我正在使用的包装器的测试:

class ApplicationSpec extends Specification {

  implicit object PositionFormat extends Format[List[List[Position]]] {
    def writes(historyList: List[List[Position]]) : JsValue = {
      Json.arr(historyList.map{
        o => Json.arr(o.map{ p =>
          Json.obj(
            "amount"    -> JsNumber(p.amount),
            "minAmount" -> JsNumber(p.minAmount),
            "price"     -> JsNumber(p.price),
            "volume"    -> JsNumber(p.volume)
          )
        })
      })
    }
    def reads(json: JsValue): JsResult[List[List[Position]]] = ???
  }

  "Application" should {

    "Convert position data to json" in {
      val l1 = ListBuffer(new Position(1.0D,2.0D,3.0D,4.0D),
        new Position(5.0D,6.0D,7.0D,8.0D),
        new Position(9.0D,10.0D,11.0D,12.0D)).toList
      val l2 = ListBuffer(new Position(0.1D,0.2D,0.3D,0.4D),
        new Position(5.0D,6.0D,7.0D,8.0D),
        new Position(9.0D,10.0D,11.0D,12.0D)).toList
      val obj = ListBuffer(l1,l2).toList
      val json = Json.toJson(obj)
      var string: String = json.toString()
      println(string)
    }
  }
}
4

1 回答 1

9

似乎Json.arr接受它的参数并JsValue为它们的 JSON 数组返回一个。看起来您可以简单地使用Json.toJson

以下arr是使用方法:

// Json.arr
Json.arr("one", "two")
// Gives you
// play.api.libs.json.JsArray = ["one","two"]

如果您改为:

// vs
val l = List("one", "two")
Json.arr(l)
// Gives you
// play.api.libs.json.JsArray = [["one","two"]]
// ... a nested array, which is what you don't want.

你需要的是:

// Json.toJson
Json.toJson(l)
// Gives you:
// play.api.libs.json.JsValue = ["one","two"]
于 2013-04-15T04:42:07.437 回答