在java中使用jackson json反序列化时忽略缺少的属性 [英] Ignore missing properties during jackson json deserialize in java

问题描述

在示例中

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);

它会抛出一个JsonMappingException，说它不能反序列化。是否有注释在反序列化期间忽略缺失的字段？

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

谢谢

推荐答案

我认为你想要的是

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

这是Jackson 1.x的方式。我认为在2.x中有一种新的方式。类似

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只序列化非空值，不要抱怨反序列化缺失值时。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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