playframework json 写问题

Question

case class CustomerInfo (
    customerId: Long,
    customerName: String,
    cameAt: String,
    staffId:String,
    staffName:String
)

case class CustomerResult (
    customers:Option[Seq[CustomerInfo]] = None
){
    def toJson = Json.toJson(this)
}

implicit val customerInfoWrites: Writes[CustomerInfo] = (
        (JsPath \ "customer_id").write[Long] and
      (JsPath \ "customer_name").write[String] and
      (JsPath \ "came_at").write[String] and
      (JsPath \ "staff_id").write[String] and
        (JsPath \ "staff_name").write[String]
)(unlift(CustomerInfo.unapply))

implicit val custmerResultWrites: Writes[CustomerResult] = (
      (JsPath \ "customers").writeNullable[Seq[CustomerInfo]]
)(unlift(CustomerResult.unapply))

第二种方法是错误的custmerResultWrites，因为它只有一个JsPath write.If我在CustomerResult中添加了一个部分，错误是可以的。错误：

found   : CustomersSearchResult.this.CustomerResult => Option[Seq[CustomersSearchResult.this.CustomerInfo]]
required: play.api.libs.json.Writes[Seq[CustomersSearchResult.this.CustomerInfo]]

Answer 1

一般来说，该错误是由于使用 combinators with a single field（请参阅该问题的一般解决方法的答案。）

一种方法是：

implicit val custmerResultWrites: Writes[CustomerResult] =
  (JsPath \ "customers").writeNullable[Seq[CustomerInfo]]
                        .contramap(_.customers)

如果 customers 是空的，这会给你一个空对象 ({})。相反，如果你想要一个空列表，你可以这样做：

val custmerResultWrites2: Writes[CustomerResult] =
  (JsPath \ "customers").writeNullable[Seq[CustomerInfo]].contramap {
    case CustomerResult(None) => Some(Seq.empty[CustomerInfo])
    case CustomerResult(seq) => seq
  }

这将导致：

{
  "customers" : [ ]
}

playframework json 写问题

playframework json write quesstion

scala

playframework

playframework-2.0