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Play Framework 2.3 Scala-使用隐式写入转换器将嵌套对象序列化为JSon_Json_Scala_Playframework_Jsonserializer_Playframework Json - Fatal编程技术网

Play Framework 2.3 Scala-使用隐式写入转换器将嵌套对象序列化为JSon

Play Framework 2.3 Scala-使用隐式写入转换器将嵌套对象序列化为JSon,json,scala,playframework,jsonserializer,playframework-json,Json,Scala,Playframework,Jsonserializer,Playframework Json,我需要一个前端jquery组件,一个像这样的特定json对象(ajax响应): 在后端,数据通过anorm(MySQL)读取,并在以下对象中转换: List(Map("division" -> "IT", "contacts" -> List(c3,c4)), Map("division" -> "Sales", "contacts" -> List(c3,c4))) 然后我尝试将对象序列化为JSon,但没有成功(包括隐式写入转换器)。下面我以同样的方

我需要一个前端jquery组件,一个像这样的特定json对象(ajax响应):

在后端,数据通过anorm(MySQL)读取,并在以下对象中转换:

List(Map("division" -> "IT", "contacts" -> List(c3,c4)), Map("division" -> "Sales",          "contacts" -> List(c3,c4)))
然后我尝试将对象序列化为JSon,但没有成功(包括隐式写入转换器)。下面我以同样的方式制作了一个简化的测试用例Idea工作表:

import play.api.libs.json.{JsValue, Json, Writes}

case class Contact(firstname: String, surname: String, empID: Option[Int])

case class ContactDivisionList(division: String, contacts: Seq[Contact])

implicit val ctWrites = new Writes[Contact] {
  def writes(ct: Contact) = Json.obj(
    "firstname" -> ct.firstname,
    "surname" -> ct.surname,
    "empid" -> ct.empID
  )
}

implicit val sdlWrites = new Writes[ContactDivisionList] {
  def writes(dlist: ContactDivisionList) = Json.obj(
    "division" -> dlist.division,
    "contacts" -> Json.toJson(dlist.contacts)
  )
}


/*
Example
*/
val c1 = Contact("Carl","Smith",Option(1))
val c2 = Contact("Henry","Miller",Option(2))
val c3 = Contact("Nancy","McDonald",Option(3))
val c4 = Contact("Susan","McBright",Option(4))

//Test case 1 ->OK
Json.toJson(List(c1,c2,c3,c4))

//Test case 2 ->OK
val c_comp1=List(Map("contacts" -> List(c1,c2)),Map("contacts" -> List(c3,c4)))
//RESULT --> c_comp1: List[scala.collection.immutable.Map[String,List[Contact]]] =     List(Map(contacts -> List(Contact(Carl,Smith,Some(1)), Contact(Henry,Miller,Some(2)))),   Map(contacts -> List(Contact(Nancy,McDonald,Some(3)), Contact(Susan,McBright,Some(4)))))
Json.toJson(c_comp1)
//res1: play.api.libs.json.JsValue = [{"contacts":    [{"firstname":"Carl","surname":"Smith","empid":1},{"firstname":"Henry","surname":"Miller","empid":2}]},{"contacts":[{"firstname":"Nancy","surname":"McDonald","empid":3},{"firstname":"Susan","surname":"McBright","empid":4}]}]


//Test case 3 ->Fail!!!
val c_comp2 = List(Map("division" -> "IT", "contacts" -> List(c1,c2)),Map("division" ->  "Sales", "contacts" -> List(c3,c4)))
//sdlWrites: play.api.libs.json.Writes[ContactDivisionList]{def writes(dlist:    ContactDivisionList): play.api.libs.json.JsObject} = $anon$2@3738baec

Json.toJson(c_comp2)
//!!!!!Error messages
/*Error:(39, 13) No Json serializer found for type     List[scala.collection.immutable.Map[String,java.io.Serializable]]. Try to implement an    implicit Writes or Format for this type.
Json.toJson(c_comp2)
^

Error:(39, 13) not enough arguments for method toJson: (implicit tjs:    play.api.libs.json.Writes[List[scala.collection.immutable.Map[String,java.io.Serializable]]    ])play.api.libs.json.JsValue.
Unspecified value parameter tjs.
Json.toJson(c_comp2)
^
*/
在脚本的末尾,您可以看到“测试用例3”,在我执行Json.toJson(c_comp2)-->“找不到类型为..的Json序列化程序”时出错。我尝试了很多东西,但都做得不对。成功的“测试用例2”的唯一区别是我用一个字符串元组扩展了映射

我希望sombody能帮我解决这个问题Thx

致意
Karsten

您的问题是,您拥有的
映射
字符串
映射到
字符串
(您的部门名称)和
列表[联系人]
的最小上界,这恰好是
java.io.Serializable

scala> case class Contact(firstname: String, surname: String, empID: Option[Int])
defined class Contact

scala> case class ContactDivisionList(division: String, contacts: Seq[Contact])
defined class ContactDivisionList

scala> val c1 = Contact("Carl","Smith",Option(1))
c1: Contact = Contact(Carl,Smith,Some(1))

scala> val c2 = Contact("Henry","Miller",Option(2))
c2: Contact = Contact(Henry,Miller,Some(2))

scala> val c3 = Contact("Nancy","McDonald",Option(3))
c3: Contact = Contact(Nancy,McDonald,Some(3))

scala> val c4 = Contact("Susan","McBright",Option(4))
c4: Contact = Contact(Susan,McBright,Some(4))

scala> Map("division" -> "IT", "contacts" -> List(c1,c2))
res10: scala.collection.immutable.Map[String,java.io.Serializable] = Map(division -> IT, contacts -> List(Contact(Carl,Smith,Some(1)), Contact(Henry,Miller,Some(2))))
我不完全确定问题的性质,但如果您已经有了
列表[ContactDivisionList]
,将其序列化为JSON非常简单:

scala> implicit val contactWrites = Json.writes[Contact]
contactWrites: play.api.libs.json.OWrites[Contact] = play.api.libs.json.OWrites$$anon$2@3676af92

scala> implicit val contactDivisionListWrites = Json.writes[ContactDivisionList]
contactDivisionListWrites: play.api.libs.json.OWrites[ContactDivisionList] = play.api.libs.json.OWrites$$anon$2@2999d17d

scala> Json.toJson(List(ContactDivisionList("IT", List(c1,c2)), ContactDivisionList("Sales", List(c3, c4))))
res2: play.api.libs.json.JsValue = [{"division":"IT","contacts":[{"firstname":"Carl","surname":"Smith","empID":1},{"firstname":"Henry","surname":"Miller","empID":2}]},{"division":"Sales","contacts":[{"firstname":"Nancy","surname":"McDonald","empID":3},{"firstname":"Susan","surname":"McBright","empID":4}]}]

在我看来,您首先应该避免使用
映射。我以前从未与anorm合作过,但我认为您需要关注的是数据结构的代码,因为在这一点上,您已经失去了类型安全性。理想情况下,您应该使用
ContactDivisionList
或直接使用
JsValue
案例构建对象。

谢谢jilen,但是在脚本“ContactDivisionList”中,您建议的案例类ContactGroup和我的案例类有什么不同吗?
scala> implicit val contactWrites = Json.writes[Contact]
contactWrites: play.api.libs.json.OWrites[Contact] = play.api.libs.json.OWrites$$anon$2@3676af92

scala> implicit val contactDivisionListWrites = Json.writes[ContactDivisionList]
contactDivisionListWrites: play.api.libs.json.OWrites[ContactDivisionList] = play.api.libs.json.OWrites$$anon$2@2999d17d

scala> Json.toJson(List(ContactDivisionList("IT", List(c1,c2)), ContactDivisionList("Sales", List(c3, c4))))
res2: play.api.libs.json.JsValue = [{"division":"IT","contacts":[{"firstname":"Carl","surname":"Smith","empID":1},{"firstname":"Henry","surname":"Miller","empID":2}]},{"division":"Sales","contacts":[{"firstname":"Nancy","surname":"McDonald","empID":3},{"firstname":"Susan","surname":"McBright","empID":4}]}]