JsonMapping异常无法构造实例

Question

我有一个关注class：

@Data
@NoArgsConstructor
public class FloorPriceData {
  Double multiplicationFactor;
  Double additionFactor;
  Integer heuristicValue;

  public static void main(String[] args) {
    String a = "{\"multiplicationFactor\" : 3, \"additionFactor\" : 1, \"heuristicValue\" : 3}";
    System.out.println(Utils.getMapper().convertValue(a, FloorPriceData.class));
  }
}

当我尝试转换时 JSON

{"multiplicationFactor" : 3, "additionFactor" : 1, "heuristicValue" : 3}

对于这个 class 实例，我得到以下异常：

Getting Caused by: com.fasterxml.jackson.databind.JsonMappingException: Can not construct instance of com.medianet.rtb.mowgli.commons.dto.adexchange.FloorPriceData: no String-argument constructor/factory method to deserialize from String value ('{"multiplicationFactor" : 3, "additionFactor" : 1, "heuristicValue" : 3}')

Answer 1

在这种情况下，您应该使用 ObjectMapper.readValue(String json, Class<T> valueType):

System.out.println(Utils.getMapper().readValue(a, FloorPriceData.class));

它将生成一个输出：

FloorPriceData(multiplicationFactor=3.0, additionFactor=1.0, heuristicValue=3)

当您尝试反序列化 JSON 以反对

Utils.getMapper().convertValue(a, FloorPriceData.class)

失败，因为convertValue首先序列化给定值，然后再次反序列化：

This is functionality equivalent to first serializing given value into JSON, then binding JSON data into value of given type, but may be executed without fully serializing into JSON.

在这种情况下，它花费了：

{"multiplicationFactor" : 3, "additionFactor" : 1, "heuristicValue" : 3}

并将其序列化为：

"{\"multiplicationFactor\" : 3, \"additionFactor\" : 1, \"heuristicValue\" : 3}"

JsonMapping异常无法构造实例

JsonMapping exception Cannot construct instance

java

jackson

objectmapper