Question

我正在尝试将列表列表转换为Json。到目前为止，我能够将第一个项目转换为Json，但是如何为整个列表执行此操作？

目前我有以下内容：

  def jsonAll = DBAction { implicit rs =>
    val list = Performances.listAll

    val w = Json.obj(
      "items" -> Json.arr(
        Json.obj("performance" -> list(0)._1),
        Json.obj("location" -> list(0)._2),
        Json.obj("user" -> list(0)._3)
      )
    )

    Ok(Json.toJson(w))
  }

'list'定义为：

  def listAll: List[(Performance, Location, DBUser)] = {
    ...
  }

注意：您还必须定义单个对象的格式。

  implicit val performanceFormat = Json.format[Performance]
  implicit val locationFormat = Json.format[Location]
  implicit val userFormat = Json.format[DBUser]

Answer 1

您是否考虑过使用自己的演出类型并使用Json.toJson？

case class Performance(performance: String, location: String, user: String)

implicit val performancesWrites = new Writes[Performance] {
  def writes(p: Performance) = Json.obj(
    "performance" -> p.performance,
    "location" -> p.location,
    "user" -> p.user
  )
}

val list = List(("hi", "there", "buddy"), ("meep", "eleven", "foobar"))
val performances = list.map((Performance.apply _) tupled)    
val resultingJson = Json.toJson(performances)

// Yields: [{"performance":"hi","location":"there","user":"buddy"},{"performance":"meep","location":"eleven","user":"foobar"}]

Answer 2

所以更多的研究和潜入Scala的奇怪世界（我来自Java）我找到了解决方案：

  def jsonAll = DBAction { implicit rs =>
    val list = Performances.listAll

    val re = list.map(
      iter =>
        Json.obj(
          "performance" -> iter._1,
          "location" -> iter._2,
          "user" -> iter._3
        )
    )

    Ok(Json.toJson(re))
  }

一旦你知道语法就很简单：|

Scala Play将List列表转换为Json

2 个答案: