2017-08-24 51 views
1

获得对象的名单我想使用akka-http-spray-json 10.0.9spray-json。如何从JSON

我的模型:

case class Person(id: Long, name: String, age: Int) 

我得到JSON字符串jsonStr与人的名单,并尝试分析它:

implicit val personFormat: RootJsonFormat[Person] = jsonFormat3(Person) 

val json = jsonStr.parseJson 
val persons = json.convertTo[Seq[Person]] 

错误:

Object expected in field 'id'


可能是我需要创建implicit object extends RootJsonFormat[List[Person]]并覆盖readwrite方法。

implicit object personsListFormat extends RootJsonFormat[List[Person]] { 
    override def write(persons: List[Person]) = ??? 

    override def read(json: JsValue) = { 
     // Maybe something like 
     // json.map(_.convertTo[Person]) 
     // But there is no map or similar method :(
    } 
} 

P.S.对不起,我的英语不是我的母语。

UPD
jsonStr:

[ {"id":6,"name":"Martin Ordersky","age":50}, {"id":8,"name":"Linus Torwalds","age":43}, {"id":9,"name":"James Gosling","age":45}, {"id":10,"name":"Bjarne Stroustrup","age":59} ] 

回答

1

我得到完全预期的结果有:

import spray.json._ 

object MyJsonProtocol extends DefaultJsonProtocol { 
    implicit val personFormat: JsonFormat[Person] = jsonFormat3(Person) 
} 

import MyJsonProtocol._ 

val jsonStr = """[{"id":1,"name":"john","age":40}]""" 
val json = jsonStr.parseJson 
val persons = json.convertTo[List[Person]] 
persons.foreach(println) 
+0

我得到了'找不到类型CustomCollectionFormats' – Oleg

+0

对不起..这来自于我们自己的lib .. –

+0

对不起,编辑和不相关的评论..我的scala repl命名空间充满了左转.. –