Java杰克逊：JSON值未知

Question

我还在学习如何使用Jackson。。。

所以我有一个JSON对象，它的值有时是整数、长字符串或列表

值：整数

{
  "id":1,
  "active":1,
  "name":"name1",
  "value":155,
  ...

值：字符串

{
  "id":2,
  "active":1,
  "name":"name2",
  "value":"Lorem Ipsum is simply dummy text of the printing and typesetting industry. Lorem Ipsum has been the industry's standard dummy text ever since the 1500s, when an unknown printer took a galley of type and scrambled it to make a type specimen book...",
  ...

价值：列表

{
  "id":3,
  "active":1,
  "name":"name3",
  "value":[
    "One",
    "Two",
    "Three",
    "Four"],
  ...

所以总的来说。。。

{
  {
      "id":1,
      "active":1,
      "name":"name1",
      "value":155,
      ...
  },
  {
      "id":2,
      "active":1,
      "name":"name2",
      "value":"Lorem Ipsum is simply dummy text of the printing and typesetting industry. Lorem Ipsum has been the industry's standard dummy text ever since the 1500s, when an unknown printer took a galley of type and scrambled it to make a type specimen book...",
      ...
  },
  {
      "id":3,
      "active":1,
      "name":"name3",
      "value":[
        "One",
        "Two",
        "Three",
        "Four"],
      ...
  }
}

这是我的POJO模型

@JsonIgnoreProperties(ignoreUnknown=true)
@JsonAutoDetect(fieldVisibility= JsonAutoDetect.Visibility.ANY)
public class OQScoresRows {
  private int id;
  private int active;
  private String name;
  private List<String> value;
  ... ...

这是我的映射器代码

ObjectMapper mapper = new ObjectMapper();
try{
  POJO obj = mapper.readValue(<JSONOBJECT>, POJO.class);
}catch(JsonParseException e){
  return mapper.writeValueAsString(e);
}

问题是，当我执行代码时，出现以下错误：

com.fasterxml.jackson.databind.JsonMappingException: Can not deserialize instance of java.util.ArrayList out of VALUE_NUMBER_INT token

我很清楚，之所以会发生这种情况，是因为“value”可以包含三种不同类型中的一种，我如何使代码足够灵活以适应这些类型。。。我总是可以在其他方法中检测值是int、List还是String，但我首先需要建模（不是我）。。。

我的问题很简单：如何使我的代码足够灵活以适应类型。。。

Answer 1

If-else语句可以回答您的问题，但它在json对象中都是字符串格式，所以您必须找出一种从值中标识数据类型的方法。例如，<代码>整数。valueOf（value）到identity int；从标识列表开始；另一个是字符串类型。您可以参考此答案，这是将json字符串转换为对象或列表的一般方法。

Answer 2

如果它可以是Intger、List和String中的任何一个，那么您可以将其声明为Object并稍后使用instanceof对其进行强制转换，例如：

@JsonIgnoreProperties(ignoreUnknown=true)
@JsonAutoDetect(fieldVisibility= JsonAutoDetect.Visibility.ANY)
public class OQScoresRows {
  private int id;
  private int active;
  private String name;
  private Object value;

在反序列化之后，可以编写与以下类似的逻辑：

if(value instanceof Integer){
    //do something after casting it to Integer
}else if(value instanceof List){
    //do something after casting it to List
}else if(value instanceof String){
    do something after casting it to String
}