在 Java 中的 Jackson JSON 反序列化期间忽略缺少的属性 [英] Ignore missing properties during Jackson JSON deserialization in Java

问题描述

示例中

class Person {
   String name;
   int age;
}

如果 JSON 对象缺少属性 'age'，

If the JSON object has a missing property 'age',

{
    "name": "John"
}

Person person = objectMapper.readValue(jsonFileReader, Person.class);

它抛出一个 JsonMappingException 说它不能反序列化.是否有注释可以在反序列化过程中忽略缺失的字段?

it throws a JsonMappingException saying it cannot deserialize. Is there an annotation to ignore missing fields during deserialization?

推荐答案

我想你想要的是

@JsonSerialize(include = JsonSerialize.Inclusion.NON_NULL)
public class Person {
  ...
}

这就是 Jackson 1.x 的方式.我认为 2.x 中有一种新方法.类似的东西

that's the Jackson 1.x way. I think there's a new way in 2.x. Something like

@JsonInclude(Include.NON_NULL)
public class Person {
  ...
}

这些将告诉 Jackson 仅序列化非空值，并且在反序列化缺失值时不要抱怨.我认为它只会将其设置为 Java 默认值.

These will tell Jackson to only serialize values that are not null, and don't complain when deserializing a missing value. I think it will just set it to the Java default.

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