创建读取的列表[案例类]

Question

我有3个case类：

scala> case class Friend(id: Long, hobbies: List[Long]) 
defined class Friend 

scala> case class Kid(friends: List[Friend]) 
defined class Kid 

scala> case class Parent(kids: List[Kid]) 
defined class Parent 

它们可以显示在层次结构（父母有孩子（一个或多个）的朋友（S）有（ID和爱好）。

Parent 
    ---> Kid 
     ---> Friend 
      ---> id: Long 
      ---> hobbies: List[Long] 

我创建了FriendsReads（尽管作为senia指出here，我可能只需要使用

scala> implicit val FriendReads: Reads[Friend] = Json.format[Friend] 
FriendReads: play.api.libs.json.Reads[Friend] = play.api.libs.json.OFormat$$anon$1@236b2692 

scala> implicit val FriendReads: Reads[Friend] = (
    | (JsPath \ "id").read[Long] and 
    | (JsPath \ "hobbies").read[List[Long]] 
    |)(Friend.apply _) 
FriendReads: play.api.libs.json.Reads[Friend] = play.api.libs.json.Reads$$anon$8@4bb7bb4 

然后，当我试图创建KidReads时，我遇到了编译时问题。

scala> implicit val KidReads: Reads[Kid] = (
    | (JsPath \ "friends").read[List[Friend]] 
    |)(Kid.apply _) 
<console>:40: error: overloaded method value read with alternatives: 
    (t: List[Friend])play.api.libs.json.Reads[List[Friend]] <and> 
    (implicit r: play.api.libs.json.Reads[List[Friend]])play.api.libs.json.Reads[List[Friend]] 
cannot be applied to (List[Friend] => Kid) 
     (JsPath \ "friends").read[List[Friend]] 
          ^

如何解决此错误？

Answer 1

表达(JsPath \ "friends").read[List[Friend]]创建Reads[List[Friend]]，你可以使用map方法这样将其转换为Reads[Kid]：

implicit val KidReads: Reads[Kid] = 
    (JsPath \ "friends").read[List[Friend]] map Kid.apply 

功能Kid.apply将被应用到结果的Reads[List[Friend]]#reads这样的：

val friendListReads: Reads[List[Friend]] = (JsPath \ "friends").read[List[Friend]] 

implicit val kidReads: Reads[Kid] = new Reads[Kid] { 
    def reads(json: JsValue): JsResult[Kid] = friendListReads.reads(json) match { 
    case JsSuccess(fList, path) = JsSuccess(Kid(fList), path) 
    case e: JsError => e 
    } 
}