為Scup的Seq播放Scala Json作家

Question

我正在嘗試找到一種方法來使用內置的Macro Json Writer來序列化Seq [（String，Customer）]

我設法為Seq [Customer]做了這個，但是當添加touple時，編譯器開始尖叫我。

此代碼有效：

package models.health

import play.api.libs.json._

case class Customer(name: String, age: Int)

//we use the dummy var as a workaround to the json writer    limitations (cannot handle single argument case class)
case class Demo(customers: Seq[Customer], dummy: Option[String] =    None)

object Demo {

 import play.api.libs.functional.syntax._

 implicit val customer_writer = Json.writes[Customer]

 implicit val writes: Writes[Demo] = (
 (__ \ "customers").write[Seq[Customer]] and
 (__ \ "dummy").writeNullable[String]) { 
    (d: Demo) => (d.customers,d.dummy)
 }

}

但是下面的代碼（只需從Seq [Customer]更改為Seq [（String，Customer）]不會Copmile ...真的很感激任何幫助：

package models.health

import play.api.libs.json._

case class Customer(name: String, age: Int)

//we use the dummy var as a workaround to the json writer    limitations (cannot handle single argument case class)
case class Demo(customers: Seq[(String,Customer], dummy: Option[String] =    None)

object Demo {

 import play.api.libs.functional.syntax._

 implicit val customer_writer = Json.writes[Customer]

 implicit val writes: Writes[Demo] = (
 (__ \ "customers").write[Seq[(String,Customer)]] and
 (__ \ "dummy").writeNullable[String]) { 
    (d: Demo) => (d.customers,d.dummy)
 }

}

這是我得到的編譯器錯誤：

No Json serializer found for type Seq[(String,models.health.Customer)]

Answer 1

庫不會假設您希望元組如何序列化。 你可以使用數組，對象等。

通過添加此隱式Writes函數，您的序列化程序將其作為數組寫出。

  implicit def tuple2Writes[A, B](implicit a: Writes[A], b: Writes[B]): Writes[Tuple2[A, B]] = new Writes[Tuple2[A, B]] {
    def writes(tuple: Tuple2[A, B]) = JsArray(Seq(a.writes(tuple._1), b.writes(tuple._2)))
  }