Spring boot jackson-使用根名反序列化Json

Question

{
    "user": {
        "name": "Ram",
        "age": 27
    }
}

public class User {
    private String name;
    private int age;

    // getters & setters
}

@Configuration
public class JacksonConfig {

    @Bean
    public Jackson2ObjectMapperBuilder jacksonBuilder() {
        Jackson2ObjectMapperBuilder builder = new Jackson2ObjectMapperBuilder();
        builder.featuresToEnable(DeserializationFeature.UNWRAP_ROOT_VALUE);
        return builder;
    }
}

{
 "name": "Ram",
 "age": 27
}

我想用Spring靴的方式来做这件事。

Answer 1

使用ObjectMapper可以很容易地解决这个问题。下面是您必须做的：-如下所示注释User类

@JsonRootName("user")
    public class User {
    private String name;
    private int age;

        // getters & setters
    }

>

创建CustomJsonMapper类

public class CustomJsonMapper extends ObjectMapper {

    private DeserializationFeature deserializationFeature;

    public void setDeserializationFeature (DeserializationFeature  deserializationFeature) {
        this.deserializationFeature = deserializationFeature;
        enable(this.deserializationFeature);
    }
}

等效Spring配置

<bean id="objectMapper" class=" com.cognizant.tranzform.notification.constant.CustomJsonMapper">
    <property name="deserializationFeature" ref="deserializationFeature"/>
</bean>
<bean id="deserializationFeature" class="com.fasterxml.jackson.databind.DeserializationFeature"
      factory-method="valueOf">
    <constructor-arg>
        <value>UNWRAP_ROOT_VALUE</value>
    </constructor-arg>
</bean>

ApplicationContext context = new ClassPathXmlApplicationContext(
            "applicationContext.xml");
    ObjectMapper objectMapper = (ObjectMapper) context
            .getBean("objectMapper");
    String json = "{\"user\":{ \"name\": \"Ram\",\"age\": 27}}";
    User user = objectMapper.readValue(json, User.class);