我有一个班级
case class Token(
creationDate: Date,
expires: Option[Date]
) {
def toJson(): String = Json.stringify(Json.toJson(this))
}
object Token {
def fromJson(json: String): Token = Json.parse(json).as[Token]
implicit val tokenReads: Reads[Token] = (
(JsPath \ "creation_date").read[Date] and
(JsPath \ "expires").readNullable[Date]
) (Token.apply _)
implicit val tokenWrites: Writes[Token] = (
(JsPath \ "creation_date").write[Date] and
(JsPath \ "expires").writeNullable[Date]
) (unlift(Token.unapply))
}
这是从json创建的,如
{
"creation_date": "2014-05-22T08:05:57.556385+00:00",
"expires": null
}
问题是日期格式无法直接转换为日期,我希望以某种方式获取该字符串,并使用
转换它DateFormat df2 = new SimpleDateFormat("yyyy-MM-dd'T'HH:mm:ss.SSSXXX");
String string2 = "2001-07-04T12:08:56.235-07:00";
Date result2 = df2.parse(string2);
然后将其传递给Token构造函数,但我似乎无法弄清楚如何在apply函数中执行此操作
答案 0 :(得分:0)
如果您有特殊日期格式
,则可以将字符串映射到日期def strToDate(string2: String): Date = {
//... something such
val df2 = new SimpleDateFormat("yyyy-MM-dd'T'HH:mm:ss.SSSXXX");
df2.parse(string2);
}
implicit val tokenReads: Reads[Token] = (
(JsPath \ "creation_date").read[String].map(strToDate) and
(JsPath \ "expires").readNullable[String].map(_.map(strToDate))
) (Token.apply _)