我正在尝试在2.1RC Play Framework中将Scala转换为JSON .
我可以执行以下操作并获取JSON:
import play.api.libs.json._
val a1=Map("val1"->"a", "val2"->"b")
Json.toJSon(a1)
因为a1只是Map [String,String]才能正常工作 .
但是,如果我有更复杂的东西,比如我有Map [String,Object],那就不行了:
val a = Map("val1" -> "xxx", "val2"-> List("a", "b", "c"))
Json.toJSon(a1)
>>> error: No Json deserializer found for type scala.collection.immutable.Map[String,Object]
我发现我可以做以下事情:
val a2 = Map("val1" -> Json.toJson("a"), "val2" -> Json.toJson(List("a", "b", "c")))
Json.toJson(a2)
这很有效 .
但是我怎么能以一般的方式做到这一点?我以为我可以做以下事情:
a.map{ case(k,v)=> (k, Json.toJson(v) )}
>>> error: No Json deserializer found for type Object
但我仍然得到一个错误,它无法反序列化
Additional Information:
Json.toJson可以将Map [String,String]转换为JsValue:
scala> val b = Map( "1" -> "A", "2" -> "B", "3" -> "C", "4" -> "D" )
b: scala.collection.immutable.Map[String,String] = Map(1 -> A, 2 -> B, 3 -> C, 4 -> D)
scala> Json.toJson(b)
res31: play.api.libs.json.JsValue = {"1":"A","2":"B","3":"C","4":"D"}
但是,它无法尝试转换Map [String,Object]:
scala> a
res34: scala.collection.immutable.Map[String,Object] = Map(val1 -> xxx, val2 -> List(a, b, c))
scala> Json.toJson(a)
<console>:12: error: No Json deserializer found for type scala.collection.immutable.Map[String,Object]. Try to implement an implicit Writes or Format for this type.
Json.toJson(a)
在将Scala转换为Json时使用此Play Framework页面中的'hint',我发现以下内容(http://www.playframework.org/documentation/2.0.1/ScalaJson):
如果不是Map [String,Object],而是Map [String,JsValue],那么Json.toJson()将起作用:
scala> val c = Map("aa" -> Json.toJson("xxxx"), "bb" -> Json.toJson( List("11", "22", "33") ) )
c: scala.collection.immutable.Map[String,play.api.libs.json.JsValue] = Map(aa -> "xxxx", bb -> ["11","22","33"])
scala> Json.toJson(c)
res36: play.api.libs.json.JsValue = {"aa":"xxxx","bb":["11","22","33"]}
所以,我想要的是,给定一个Map [String,Object],我知道Object值最初都是String或List [String]类型,如何将函数Json.toJson()应用于all map中的值并获取Map [String,JsValue] .
我还发现我可以过滤掉那些纯粹是字符串的值和那些类型为List [String]的值:
scala> val a1 = a.filter({case(k,v) => v.isInstanceOf[String]})
a1: scala.collection.immutable.Map[String,Object] = Map(val1 -> xxx)
scala> val a2 = a.filter({case(k,v) => v.isInstanceOf[List[String]]})
<console>:11: warning: non-variable type argument String in type List[String] is unchecked since it is eliminated by erasure
val a2 = a.filter({case(k,v) => v.isInstanceOf[List[String]]})
^
a2: scala.collection.immutable.Map[String,Object] = Map(val2 -> List(a, b, c))
List [String]过滤会发出警告,但似乎给出了我想要的答案 . 如果可以应用两个过滤器,然后在结果的值上使用Json.toJson(),并且结果合并,也许这会起作用?
但过滤结果仍然是Map [String,Object]类型,这会导致问题:
scala> Json.toJson(a1)
<console>:13: error: No Json deserializer found for type scala.collection.immutable.Map[String,Object]. Try to implement an implicit Writes or Format for this type.
Json.toJson(a1)
1 回答
Play 2.1 JSON API不为Type
Map[String, Ojbect]
提供序列化程序 .为特定类型定义
case class
和Format
而不是Map[String, Object]
:如果您不想创建案例类 . 以下代码为Map [String,Object]提供了JSON序列化器/反序列化器:
更动态
示例代码:
输出: