在scala中使用递归函数解析复杂地图

Question

我有一个像下面这样的 json 文件

{

"Item Version": 1.0,
"Action2": "true12",
"Item Creation Time": "2019-04-14 14:15:09",
"Trade Dictionary": {
    "Country": "India",
    "TradeNumber": "1",
    "action": {
        "Action2": true,
        "Action1": false
    },
    "Value": "XXXXXXXXXXXXXXX"
},
"Payments": {
    "abc": "def",
    "Payment Details": [{
        "Payment Date": "2019-04-11",
        "Payment Type": [{
            "Payval": "myval "
        }]
    }]
}

}

我有一个这样的 scala 代码

def secondFlatten(map: Map[String, Any]):Map[String, Any]={val c=map.flatten {

val c=map.flatten {

 case ((key, map : Map[String, Any])) => {
   map.flatten {
     case ((innerKey: String, l: List[Map[String, Any]])) => l.head.flatten
     {

               case ((key1: String, l1: List[Map[String, Any]]))=>l1.head.map{case (x:String, y: Any)=> (s"$key->$innerKey->$key1->$x"-> y)}
               case ((key1: String, m1 : Map[String, Any])) =>  m1.map{case (x:String, y: Any)=> (s"$key->$innerKey->$key1->$x"-> y)}
               case (key1: String, value1: String) => Map(s"$key->$innerKey->$key1"-> value1)
     }
     case ((innerKey: String, m : Map[String, Any])) => m.map{case (x:String, y: Any)=> (s"$key->$innerKey->$x"-> y)}
     case (innerKey: String, value: String) => Map(s"$key->$innerKey"-> value)
}
}
case ((key, value)) => Map(key -> value)

}.toMap
return c

}

返回这样的值

（付款->付款明细->付款日期，2019-04-11）

(贸易字典->价值,XXXXXXXXXXXXXXX)

(付款->付款明细->付款类型->Payval,myval)

（贸易词典->国家，印度）

(贸易字典->贸易编号,1)

(交易字典->action->Action1,false)

(Action2,true12)

（项目版本，1.0）

(交易字典->action->Action2,true)

（物品创建时间，2019-04-14 14:15:09）

(付款->abc,def)

有没有办法写一个递归函数而不是放同一种case语句？？？

Answer 1

给你：

def flattenMap(m: Map[String, Any], tree: List[String] = List()) : Iterable[(String, Any)] = m.flatten {
  case (k: String, v: Map[String, Any] @unchecked) => flattenMap(v, tree :+ k)
  case (k: String, v: List[Map[String, Any]] @unchecked) => v.flatten(flattenMap(_, tree :+ k))
  case (k: String, v: Any) => List((tree :+ k.toString).mkString("->") -> v)
  case (k: String, null) => List((tree :+ k).mkString("->") -> null)
}

（此处回复：https://scastie.scala-lang.org/benrobby/9fWtDyvyTt28BEul6UJ4Bg）