Fre*_*ind 2 json playframework-2.0
这是我的代码:
import play.api.mvc._
import play.api.libs.json._
import play.api.libs.json.Json._
import play.api.libs.json.Writes._
class BaseController extends Controller with Secured with DefaultWrites {
private implicit def str2json(str: String) = new {
def asSuccessJson = toJson(Map("success" -> true, "message" -> str)) // (*)
def asFailedJson = toJson(Map("success" -> false, "message" -> str)) // (*)
}
}
Run Code Online (Sandbox Code Playgroud)
但它无法在两(*)行编译.错误消息是:
Multiple markers at this line
- No Json deserializer found for type scala.collection.immutable.Map[java.lang.String,Any]. Try
to implement an implicit Writes or Format for this type.
- not enough arguments for method toJson: (implicit tjs:
play.api.libs.json.Writes[scala.collection.immutable.Map[java.lang.String,Any]])
play.api.libs.json.JsValue.Unspecified value parameter tjs.
Run Code Online (Sandbox Code Playgroud)
我必须把它写成:
def asSuccessJson = toJson(Map("success" -> true.toString, "message" -> str))
Run Code Online (Sandbox Code Playgroud)
注意事项true.toString.它很有效但很无聊.
怎么解决?
这很合乎逻辑:您尝试将异构映射转换为JsValue:
Map("success" -> true, "message" -> str)是一个Map[String, Any].
没有隐式作者能够将a转换Map[String, Any]为a JsValue(并且不能有任何).
当你编写时Map("success" -> true.toString, "message" -> str),你创建一个Map [String,String],并且有一个编写器.
我会写:
def asSuccessJson = JsObject(Seq("success" -> JsBoolean(true), "message" -> JsString(str))) // (*)
Run Code Online (Sandbox Code Playgroud)
顺便说一句,JSON API肯定会在Play 2的下一个版本中"美化"一下......
| 归档时间: |
|
| 查看次数: |
6693 次 |
| 最近记录: |