如何根据条件过滤 JSON 数组结果？

Question

我正在使用 Scala Play 2.7.2 并已阅读 ScalaJsonTransformers 和 ScalaJson。调用 JSON API 后，我得到如下（简化的 MCVE）结果：

{
  "type": "searchset",
  "total": 5,
  "entry": [
    {
        "start": "2019-06-07T09:00:00",
        "end": "2019-06-07T11:00:00",
        "id": "55"
    },
    {
        "start": "2019-06-07T13:00:00",
        "end": "2019-06-07T15:00:00",
        "id": "56"
    },
    {
        "start": "2019-06-07T16:00:00",
        "end": "2019-06-07T17:00:00",
        "id": "60"
    },
    {
        "start": "2019-06-10T11:00:00",
        "end": "2019-06-10T12:00:00",
        "id": "58"
    },
    {
        "start": "2019-06-11T14:00:00",
        "end": "2019-06-11T15:00:00",
        "id": "61"
    }
  ]
}

我想过滤结果并仅选择满足条件的结果，例如，过滤掉结束日期大于某个日期val to = new DateTime("2019-06-10T00:00:00")的结果并做一些事情：

(json \\ "end").filter(new DateTime(_).isBefore(to.toDate.getTime))...

但这不起作用，因为结果是选择而不是整个 json 节点，而且它也离开了外部。

解决方案应该输出结果：

{
  "type": "searchset",
  "total": 3,
  "entry": [
    {
        "start": "2019-06-07T09:00:00",
        "end": "2019-06-07T11:00:00",
        "id": "55"
    },
    {
        "start": "2019-06-07T13:00:00",
        "end": "2019-06-07T15:00:00",
        "id": "56"
    },
    {
        "start": "2019-06-07T16:00:00",
        "end": "2019-06-07T17:00:00",
        "id": "60"
    }
}

如何使用 Play JSON 做到这一点？

Answer 1

对于从海岸到海岸的设计，尝试像这样定义update transformer

val to = new DateTime("2019-06-10T00:00:00")

val endDateFilterTransformer = (__ \ 'entry).json.update(__.read[JsArray].map {
  case JsArray(values) => JsArray(values.filter(v => (v \ "end").as[DateTime].isBefore(to)))
})

val outJson = json.transform(endDateFilterTransformer)
println(outJson.get)

哪个输出

{
  "entry": [
    {
      "start": "2019-06-07T09:00:00",
      "end": "2019-06-07T11:00:00",
      "id": "55"
    },
    {
      "start": "2019-06-07T13:00:00",
      "end": "2019-06-07T15:00:00",
      "id": "56"
    },
    {
      "start": "2019-06-07T16:00:00",
      "end": "2019-06-07T17:00:00",
      "id": "60"
    }
  ],
  "total": 5,
  "type": "searchset"
}

对于 JSON 到 OO 的设计，也可以尝试反序列化为模型

case class Entry(start: DateTime, end: DateTime, id: String)
object Entry {
  implicit val format = Json.format[Entry]
}
case class Record(`type`: String, total: Int, entry: List[Entry])

object Record {
  implicit val format = Json.format[Record]
}

然后使用常规的 Scala 方法进行过滤

val to = new DateTime("2019-06-10T00:00:00")
val record = Json.parse(raw).as[Record]
val filteredRecord = record.copy(entry = record.entry.filter(_.end.isBefore(to)))

然后像这样反序列化回 json：

Json.toJson(filteredRecord)

哪个输出

{
  "type": "searchset",
  "total": 5,
  "entry": [
    {
      "start": "2019-06-07T09:00:00.000+01:00",
      "end": "2019-06-07T11:00:00.000+01:00",
      "id": "55"
    },
    {
      "start": "2019-06-07T13:00:00.000+01:00",
      "end": "2019-06-07T15:00:00.000+01:00",
      "id": "56"
    },
    {
      "start": "2019-06-07T16:00:00.000+01:00",
      "end": "2019-06-07T17:00:00.000+01:00",
      "id": "60"
    }
  ]
}

我们使用play-json-joda 进行DateTime 序列化

libraryDependencies += "com.typesafe.play" %% "play-json-joda" % "2.7.3"
import play.api.libs.json.JodaWrites._
import play.api.libs.json.JodaReads._