Question

我有以下json输入，试图按顺序解析名称字段

scala> result
res6: play.api.libs.json.JsValue = {"L0": 
{"name":"FASHION","id":"50000"},"L1":{"name":"ACCESSORIES AND TRAVEL","id":"51000"},"L2":{"name":"FASHION ACCESSORIES","id":"51001"},"L3":{"name":"MENS FASHION ACCESSORIES","id":"51100"},"L4":{"name":"MENS HATS","id":"51204"}}

scala> result \\ "name"
res5: Seq[play.api.libs.json.JsValue] = List("ACCESSORIES AND TRAVEL", "MENS HATS", "MENS FASHION ACCESSORIES", "FASHION ACCESSORIES", "FASHION")

我正在尝试按顺序排列这些名称

List("FASHION", "ACCESSORIES AND TRAVEL", "FASHION ACCESSORIES", "MENS FASHION ACCESSORIES", "MENS HATS")

是否可以通过播放Json库来实现？

Answer 1

在Play JSON中，我始终使用case classes。因此，您的示例如下所示：

导入play.api.libs.json ._

val json = """{"L0": 
{"name":"FASHION","id":"50000"},"L1":{"name":"ACCESSORIES AND TRAVEL","id":"51000"},"L2":{"name":"FASHION ACCESSORIES","id":"51001"},"L3":{"name":"MENS FASHION ACCESSORIES","id":"51100"},"L4":{"name":"MENS HATS","id":"51204"}}
"""

case class Element(id: String, name: String)
object Element {
  implicit val jsonFormat: Format[Element] = Json.format[Element]
}

Json.parse(json).validate[Map[String, Element]] match {
  case JsSuccess(elems, _) => println(elems.toList.sortBy(_._1).map(e => e._2.name))
  case other => println(s"Handle exception $other")
}

这给您带来的好处是，您可以按键（解决方案中丢失的信息）对结果进行排序。

Scala按顺序解析json对象

1 个答案: