9

我正在尝试使用 Lift 框架反序列化 JSON 文本，但它们似乎不支持 Seq 特征（尽管支持 List ）。举个例子...

一些代表员工的 JSON 数据（带有名字和姓氏）......

{"employees":[{"fname":"Bob","lname":"Hope"},{"fname":"Bob","lname":"Smith"}]}

这是员工域对象：

case class Employee(fname: String, lname: String) { }
case class Employees(employees: Seq[Employee]) { }

这是我的 JSON 反序列化代码......

class EmployeeTest { 

  @Test def test() {
     val jsonText: String = ....
     val e = deserialize(jsonText)
  }

  def deserialize(in: String): Employees = {
    implicit val formats = net.liftweb.json.DefaultFormats
    net.liftweb.json.Serialization.read[Employees](in)
  }
}

如果我将员工域对象更改为使用 List 而不是 Seq，那么它可以工作。但如果可以的话，我真的很想使用 Seq。

这是我在运行上述代码（使用 Seq）时看到的异常：我能做些什么来让它工作吗？谢谢你的帮助！

net.liftweb.json.MappingException: unknown error
    at net.liftweb.json.Extraction$.extract(Extraction.scala:43)
    at net.liftweb.json.JsonAST$JValue.extract(JsonAST.scala:288)
    at net.liftweb.json.Serialization$.read(Serialization.scala:50)
    at EmployeeTest.deserialize(EmployeeTest.scala:20)   
    at EmployeeTest.test(EmployeeTest.scala:13)
Caused by: java.lang.UnsupportedOperationException: tail of empty list
    at scala.collection.immutable.Nil$.tail(List.scala:388)
    at scala.collection.immutable.Nil$.tail(List.scala:383)
    at net.liftweb.json.Meta$Constructor.bestMatching(Meta.scala:60)
    at net.liftweb.json.Extraction$.findBestConstructor$1(Extraction.scala:187)
    at net.liftweb.json.Extraction$.instantiate$1(Extraction.scala:192)
    at net.liftweb.json.Extraction$.newInstance$1(Extraction.scala:222)
    at net.liftweb.json.Extraction$.build$1(Extraction.scala:240)
    at net.liftweb.json.Extraction$.mkValue$1(Extraction.scala:269)
    at net.liftweb.json.Extraction$.build$1(Extraction.scala:242)
    at net.liftweb.json.Extraction$$anonfun$4.apply(Extraction.scala:194)
    at net.liftweb.json.Extraction$$anonfun$4.apply(Extraction.scala:194)
    at scala.collection.TraversableLike$$anonfun$map$1.apply(TraversableLike.scala:206)
    at scala.collection.TraversableLike$$anonfun$map$1.apply(TraversableLike.scala:206)
    at scala.collection.LinearSeqOptimized$class.foreach(LinearSeqOptimized.scala:61)
    at scala.collection.immutable.List.foreach(List.scala:45)
    at scala.collection.TraversableLike$class.map(TraversableLike.scala:206)
    at scala.collection.immutable.List.map(List.scala:45)
    at net.liftweb.json.Extraction$.instantiate$1(Extraction.scala:194)
    at net.liftweb.json.Extraction$.newInstance$1(Extraction.scala:222)
    at net.liftweb.json.Extraction$.build$1(Extraction.scala:240)
    at net.liftweb.json.Extraction$.extract(Extraction.scala:284)
    at net.liftweb.json.Extraction$.extract0(Extraction.scala:172)
    at net.liftweb.json.Extraction$.extract(Extraction.scala:40)
    ... 33 more
4

1 回答 1

13

序列化不支持 Seq，因为它不是具体类型。在反序列化期间，没有可用于决定具体实现的类型信息。例如，我们可以使用 List 作为默认实现，但以下属性将不再适用于所有类型：

deserialize(serialize(x)) == x

这个具体案例可以反序列化如下：

import net.liftweb.json._
import net.liftweb.json.JsonAST._

case class Employee(fname: String, lname: String)
case class Employees(employees: Seq[Employee])

object Test extends Application {
  implicit val formats = DefaultFormats
  val s = """ {"employees":[{"fname":"Bob","lname":"Hope"},{"fname":"Bob","lname":"Smith"}]} """

  val json = JsonParser.parse(s)
  val employees = Employees(for {
    JArray(emps) <- json \ "employees"
    emp <- emps
  } yield emp.extract[Employee])

  println(employees)
}
于 2010-11-06T22:39:50.493 回答