Question

我有一个角度应用程序将请求发送到服务器，如下所示：

$scope.downloadPartDetails = (parts, e) ->
    req = {
      method: 'POST',
      url: '/downloads/partdetails',
      headers: {
        'Content-Type': "application/json; charset=utf-8"
      },
      data: { parts: [
        {manufacturer: "mfr1", partNumber: "part num1"},
        {manufacturer: "mfr2", partNumber: "part num2"},
        {manufacturer: "mfr3", partNumber: "part num3"}
      ] }
    }
    $http(req)

它在服务器上显示如下：

JArray(List(JObject(List(JField(manufacturer,JString(mfr1)), JField(partNumber,JString(part num1)))), JObject(List(JField(manufacturer,JString(mfr2)), JField(partNumber,JString(part num2)))), JObject(List(JField(manufacturer,JString(mfr3)), JField(partNumber,JString(part num3))))))

有没有办法将其转换为List [Part]？

case class Part(mfr: String, pn: String)

Answer 1

如果在json和case类

中使用相同的字段名称会更容易

case class Part(manufacturer: String, partNumber: String)
val part: Part = jvalue.extract[Part]

如果您无法更改案例类中的字段名称，则需要按照此处的说明实现自定义序列化程序Deserialization of case object in Scala with JSON4S

Scala如何将JsonAST.JValue转换为类型？

1 个答案: