如何在circe中编码/解码json的时间戳？

Question

在光滑中使用circe来获取json中的数据时,我可以获取(Timestamp/DateTime)实体中没有日期字段的数据.但是当我Timestamp在实体中使用字段时,会抛出错误:

[error] /var/www/html/scala-api/src/main/scala/oc/api/http/routes/TestApi.scala:40: could not find implicit value for parameter encoder: io.circe.Encoder[Seq[oc.api.models.UserEntity]]
[error]             auth => complete(userDao.getAll().map(_.asJson))

这是代码,我用于Slick Entities并使用CIRCE进行json编码.

BaseTable:

abstract class BaseTable[T](tag: Tag, name: String) extends Table[T](tag, name) {
  def id = column[Long]("id", O.PrimaryKey, O.AutoInc)
  def createdAt = column[Timestamp]("created_at")
  def updatedAt = column[Timestamp]("updated_at")
  def deletedAt = column[Timestamp]("deleted_at")
}

BaseEntity:

trait BaseEntity {
  val id : Long
  def isValid : Boolean = true
}

UserEntity:createdAt生成编码器错误

case class UserEntity(id: Long, email: String, password: String, createdAt: Timestamp) extends BaseEntity

UserEntity:这非常有效

case class UserEntity(id: Long, email: String, password: String) extends BaseEntity

用户表(油滑):

object UserTables {

  class UserTable(tag : Tag) extends BaseTable[UserEntity](tag, "users") {
    def name = column[String]("name")
    def password = column[String]("password")
    def * = (id, name, password) <> (UserEntity.tupled, UserEntity.unapply)
  }

  implicit val accountsTableQ : TableQuery[UserTable] = TableQuery[UserTable]
}

我在代码中遗漏了什么吗？任何帮助将受到高度赞赏.

Answer 1

您应该在代码中使用自定义编码器和解码器,类似于:

  implicit val TimestampFormat : Encoder[Timestamp] with Decoder[Timestamp] = new Encoder[Timestamp] with Decoder[Timestamp] {
    override def apply(a: Timestamp): Json = Encoder.encodeLong.apply(a.getTime)

    override def apply(c: HCursor): Result[Timestamp] = Decoder.decodeLong.map(s => new Timestamp(s)).apply(c)
  }

将此val放在需要编码/解码时间戳的任何代码中.例如,您可以将其放在一个对象中,并在需要的地方导入该对象.