scala 如何在 Play Json 中使用 Joda DateTime
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How to use Joda DateTime with Play Json
提问by Guillaume
I'm developing a Play application, and I'm trying to use a Joda DateTime object into my case class.
我正在开发一个 Play 应用程序，我试图在我的案例类中使用一个 Joda DateTime 对象。
package model
import org.joda.time.DateTime
import play.api.libs.json._
case class User(name: String, created: DateTime)
object User {
implicit val yourJodaDateReads = Reads.jodaDateReads("yyyy-MM-dd'T'HH:mm:ss.SSSZ")
implicit val yourJodaDateWrites = Writes.jodaDateWrites("yyyy-MM-dd'T'HH:mm:ss.SSSZ'")
implicit val userFormat = Json.format[User]
def main(args: Array[String]) {
val value = Json.parse("{ \"name\" : \"hello\" , \"created\" : \"2015-07-16T20:32:04.046+02:00\" }")
println(Json.toJson(new User("user", new DateTime())))
println(Json.fromJson(value))
}
}
Based on this solution, I'm getting this error:
基于此解决方案，我收到此错误：
Error:(18, -1) Play 2 Compiler:
/activator-1.3.2/notifier-app/app/model/Test.scala:18: ambiguous implicit values:
both value yourJodaDateReads in object User of type => play.api.libs.json.Reads[org.joda.time.DateTime]
and value userFormat in object User of type => play.api.libs.json.OFormat[model.User]
I'm using Activator 1.3.2 and Play 2.3.8.
我正在使用 Activator 1.3.2 和 Play 2.3.8。
Could you please advice me ?
你能给我建议吗？
Thanks in advance.
提前致谢。
update
更新
I understand there is a conflict with the implicit value in play.api.libs.json.Reads
我知道与play.api.libs.json.Reads 中的隐含值存在冲突
implicit val DefaultJodaDateReads = jodaDateReads("yyyy-MM-dd")
How can I resolve this issue ?
我该如何解决这个问题？
回答by Guillaume
Expecting a better alternative, here my workaround:
期待更好的选择，这里是我的解决方法：
val dateFormat = "yyyy-MM-dd'T'HH:mm:ss.SSSZ"
val jodaDateReads = Reads[DateTime](js =>
js.validate[String].map[DateTime](dtString =>
DateTime.parse(dtString, DateTimeFormat.forPattern(dateFormat))
)
)
val jodaDateWrites: Writes[DateTime] = new Writes[DateTime] {
def writes(d: DateTime): JsValue = JsString(d.toString())
}
val userReads: Reads[User] = (
(JsPath \ "name").read[String] and
(JsPath \ "created").read[DateTime](jodaDateReads)
)(User.apply _)
val userWrites: Writes[User] = (
(JsPath \ "name").write[String] and
(JsPath \ "created").write[DateTime](jodaDateWrites)
)(unlift(User.unapply))
implicit val userFormat: Format[User] = Format(userReads, userWrites)
回答by Moebius
In play 2.6, the canonical way to serialize/deserialize joda DateTime json is by using the play-json-jodalibrary. Import the libraryby updating your build.sbt. Then create json reader and json writers like this :
在 play 2.6 中，序列化/反序列化 joda DateTime json 的规范方法是使用play-json-joda库。导入库的更新build.sbt。然后像这样创建 json reader 和 json writers：
import play.api.libs.json.JodaWrites
implicit val dateTimeWriter: Writes[DateTime] = JodaWrites.jodaDateWrites("dd/MM/yyyy HH:mm:ss")
import play.api.libs.json.JodaReads
implicit val dateTimeJsReader = JodaReads.jodaDateReads("yyyyMMddHHmmss")
回答by Vincil Bishop
Try these:
试试这些：
implicit val dateWrites = jodaDateWrites("yyyy-MM-dd'T'HH:mm:ss.SSSZ")
implicit val dateReads = jodaDateReads("yyyy-MM-dd'T'HH:mm:ss.SSSZ")
回答by jakehschwartz
I know this question has been answered for a while, but I found a more concise answer
我知道这个问题已经回答了一段时间，但我找到了一个更简洁的答案
val pattern = "yyyy-MM-dd'T'HH:mm:ss.SSSZ"
implicit val dateFormat = Format[DateTime](Reads.jodaDateReads(pattern), Writes.jodaDateWrites(pattern))
implicit val userFormat = Json.format[User]
回答by Evghenii Todorov
I think you should set the Usertype in Json.toJsonand Json.fromJsonfunctions. Instead of
我认为您应该设置User类型Json.toJson和Json.fromJson功能。代替
println(Json.toJson(new User("user", new DateTime())))
println(Json.fromJson(value))
try:
尝试：
println(Json.toJson[User](new User("user", new DateTime())))
println(Json.fromJson[User](value))
When you set the type explicitly framework will know what reads/writes to use.
当您显式设置类型时，框架将知道要使用的读/写。
Update:It is not necessarily to set type for Json.toJsonfunction because you pass Userobject as function argument and framework determines the type in runtime.
But for Json.fromJson[User]you must set the type, otherwise framework doesn't know type of the object you want to read.
更新：不一定要为Json.toJson函数设置类型，因为您将User对象作为函数参数传递，框架在运行时确定类型。但是因为Json.fromJson[User]您必须设置类型，否则框架不知道您要读取的对象的类型。
