Question

我正在尝试在scala中解析JSON而且我不是很成功。这是我的JSON结构：

{ 
  "data" : "test", 
  "field" : "test", 
  "listObj" : [{ 
      "data" : "testInner1", 
      "field" : "testInner1" 
   } , { 
      "data" : "testInner2", 
      "field" : "testInner2"
   }],
   "obj" : { 
      "data" : "testInner1", 
      "field" : "testInner1" 
   }
}

我正用电梯解析。这是我的代码：

import net.liftweb.json._

implicit val formats = DefaultFormats
case class MyJson(data: String, field: String, obj: MyJson, listObj: List[MyJson])
val json = parse("""
{ 
  "data" : "test", 
  "field" : "test", 
  "listObj" : [{ 
      "data" : "testInner1", 
      "field" : "testInner1" 
   } , { 
      "data" : "testInner2", 
      "field" : "testInner2"
   }],
   "obj" : { 
      "data" : "testInner1", 
      "field" : "testInner1" 
   }
}""")
println(json.extract[Query])

我是上帝以下的例外：

Caused by: net.liftweb.json.MappingException: No usable value for operation
Did not find value which can be converted into java.lang.String
at net.liftweb.json.Meta$.fail(Meta.scala:191)
at net.liftweb.json.Extraction$.mkValue$1(Extraction.scala:357)
at net.liftweb.json.Extraction$.net$liftweb$json$Extraction$$build$1(Extraction.scala:317)
at net.liftweb.json.Extraction$$anonfun$13.apply(Extraction.scala:253)
at net.liftweb.json.Extraction$$anonfun$13.apply(Extraction.scala:253)
at scala.collection.TraversableLike$$anonfun$map$1.apply(TraversableLike.scala:244)
at scala.collection.TraversableLike$$anonfun$map$1.apply(TraversableLike.scala:244)
at scala.collection.immutable.List.foreach(List.scala:318)
at scala.collection.TraversableLike$class.map(TraversableLike.scala:244)
at scala.collection.AbstractTraversable.map(Traversable.scala:105)
at net.liftweb.json.Extraction$.instantiate$1(Extraction.scala:253)
at net.liftweb.json.Extraction$.newInstance$1(Extraction.scala:286)
at net.liftweb.json.Extraction$.net$liftweb$json$Extraction$$build$1(Extraction.scala:315)
at net.liftweb.json.Extraction$.mkValue$1(Extraction.scala:351)
... 34 more
Caused by: net.liftweb.json.MappingException: Did not find value which can be converted into java.lang.String
at net.liftweb.json.Meta$.fail(Meta.scala:191)
at net.liftweb.json.Extraction$.convert(Extraction.scala:403)
at net.liftweb.json.Extraction$.net$liftweb$json$Extraction$$build$1(Extraction.scala:314)
at net.liftweb.json.Extraction$.mkValue$1(Extraction.scala:351)
... 46 more

我尝试过使用json4s，但我可以设法解析为case类。当我从我的JSON中删除obj和objList时，它运行良好。

任何人都可以指出我解析这种结构以在Scala中读取它的方法吗？

Answer 1

更改案例类定义，以便obj是可选值：

 case class MyJson(data: String, 
    field: String, 
    // obj: MyJson,        // from 
    obj: Option[MyJson],   // to
    listObj:  List[MyJson])

这是必需的，以便json提取器仍然可以生成MyJson的{{1}}实例，即使json缺少"obj"字段（它在示例中的嵌套结构中也是如此）。

listObj不需要这样做，因为json中缺少"listObj"字段仍然可以表示为List[MyJson]的空列表。

Scala Parse递归JSON

1 个答案: