Spring boot jackson - deserializes Json with root name

Question

I have the below Json

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

which I want to de-serialize into an instance of the class

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

    // getters & setters
}

For this, I have used @JsonRootName on class name and something like below

@Configuration
public class JacksonConfig {

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

But it did not work as expected. If I send something like below, it worked.

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

But I want to get the json de-serialized with root name. Can any one please suggest?

I want to spring boot way of doing this.

Possible duplicate of [Jackson JSON Deserialization with Root Element](http://stackoverflow.com/questions/11704255/jackson-json-deserialization-with-root-element) — Mifeet, May 15 '16 at 10:30
@RamBavireddi https://gist.github.com/varren/4a76ec643f5a4b26e46475a110ebe281 this definitely works for me — varren, May 15 '16 at 11:26

score 2 · Answer 1 · answered May 15 '16 at 20:13

2

@JsonRootName is a good start. Use this annotation on User class and then enable UNWRAP_ROOT_VALUE deserialization feature by adding:

spring.jackson.deserialization.UNWRAP_ROOT_VALUE=true

to your application.properties. Read more about customizing Jackson mapper in Spring Boot Reference

answered May 15 '16 at 20:13

Maciej Walkowiak

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Is it possible to do this on a per-route basis, rather than setting it for the whole application? – Dan Cancro Sep 02 '17 at 18:09
can I customize this in the project, means can i have one class to unwrap the root value and another one without unwrapping – Eldhose Jul 26 '18 at 15:46

score -1 · Answer 2 · edited Jul 25 '19 at 06:08

Using ObjectMapper you can resolve this issue easily. Here's what you have to do : - Annotate User class as given below

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

        // getters & setters
    }

Create CustomJsonMapper class

public class CustomJsonMapper extends ObjectMapper {

    private DeserializationFeature deserializationFeature;

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

Equivalent Spring configuration

<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>

Using following code you can test

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);

Can you post how you can do that using spring boot? – Ram Bavireddi May 15 '16 at 11:02 — Ram Bavireddi, May 15 '16 at 11:02

Spring boot jackson - deserializes Json with root name

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