Json输出框架/斯卡拉

Question

用框架/斯卡拉来创建Json非常简单。 这里是一个关于它的例子，

import play.api.libs.json.Json 

val json: JsValue = Json.parse(""" 
{ 
    "user": { 
    "name" : "toto", 
    "age" : 25, 
    "email" : "toto@jmail.com", 
    "isAlive" : true, 
    "friend" : { 
     "name" : "tata", 
     "age" : 20, 
     "email" : "tata@coldmail.com" 
    } 
    } 
} 
""") 

欲了解更多信息和来源：http://www.playframework.com/documentation/2.1.1/ScalaJson

Answer 1

我猜你只是想一个数组，而不是一个对象：

Json.arr(
    Json.obj(
    "id" -> "node37", 
    "name" -> 3.7, 
    "data" -> Json.obj(), 
    "children" -> Json.arr() 
), 
    Json.obj(
    "id" -> "node38", 
    "name" -> 3.8, 
    "data" -> Json.obj(), 
    "children" -> Json.arr() 
) 
) 

其中给出

[{ 
     "id": "node37", 
     "name": "3.7", 
     "data": {}, 
     "children":[] 
}, 
{ 
     "id": "node38", 
     "name": "3.8", 
     "data": {}, 
     "children":[] 
}]