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scala按顺序解析json对象

play-json scala

1

yalkris · 技术社区 · 6 年前

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")

我想把这些名字整理好

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

有没有办法用play Json库来实现这一点?

1 回复 | 直到 6 年前

1

pme 6 年前

case classes . 所以你的例子看起来像:

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")
}

这给你的是,你可以按键对结果进行排序-在你的解决方案中丢失的信息。