java - How to serialize a json containing LAZY associations

Question

I am having an Person entity which has @ManyToOne association with Contact entity with fetch type LAZY. I am using spring-boot to expose REST API. One of my POST call contains nested JSON to save the parent entity Person along with association Contact

Since Contact fetch type is LAZY, I am encountering into following exception

    ERROR 17415 --- [nio-8080-exec-4] o.a.c.c.C.[.[.[/].[dispatcherServlet] : Servlet.service() for servlet [dispatcherServlet] in context with path [] threw exception [Request processing failed; nested exception is org.springframework.http.converter.HttpMessageConversionException: Type definition error: [simple type, class org.hibernate.proxy.pojo.javassist.JavassistLazyInitializer]; nested exception is com.fasterxml.jackson.databind.exc.InvalidDefinitionException: No serializer found for class org.hibernate.proxy.pojo.javassist.JavassistLazyInitializer and no properties discovered to create BeanSerializer (to avoid exception, disable SerializationFeature.FAIL_ON_EMPTY_BEANS) (through reference chain: com.example.rest.RestResultObject["results"]->java.util.ArrayList[0]->com.example.model.Person["contact"]->com.example.model.Contact_$$_jvst8d1_4["handler"])] with root cause

com.fasterxml.jackson.databind.exc.InvalidDefinitionException: No serializer found for class org.hibernate.proxy.pojo.javassist.JavassistLazyInitializer and no properties discovered to create BeanSerializer (to avoid exception, disable SerializationFeature.FAIL_ON_EMPTY_BEANS) (through reference chain: com.example.rest.RestResultObject["results"]->java.util.ArrayList[0]->com.example.model.Person["contact"]->com.example.model.Contact_$$_jvst8d1_4["handler"])
at com.fasterxml.jackson.databind.exc.InvalidDefinitionException.from(InvalidDefinitionException.java:77) ~[jackson-databind-2.9.3.jar:2.9.3]

Without changing Contact to EAGER. Is there any best way to solve this problem?

Updated:

Person.java

public class Person {
    private long id;
    private String name;
    private String rno;
    @ManyToOne(fetch = FetchType.LAZY)
    private Contact contact;

    // Getters and setters
}

Contact.java

public class Contact {
    private long id;
    private String info;
    @OneToMany
    private List<Person> persons;
}

Answer 1

I have added the following things

• @Entity on each object
• Change long to Long in id, this will help you if you use spring data JPA
• add @Id to declare id as primary key
• @JsonBackReference & @JsonManagedReference to avoid infinite loop by Jackson

Person class

@Entity
public class Person {

 @Id
 @GeneratedValue(strategy = GenerationType.IDENTITY)
 private Long id;

 private String name;

 private String rno;

 @JsonManagedReference
 @ManyToOne(fetch = FetchType.LAZY)
 private Contact contact;

 //setter & getter

}

Contact class

@Entity
public class Contact {

 @Id
 @GeneratedValue(strategy = GenerationType.IDENTITY)
 private Long id;

 private String info;

 @JsonBackReference
 @OneToMany(cascade = CascadeType.ALL, mappedBy = "contact")
 private List<Person> persons;

 //setter & getter
}

and add a dependency

<dependency>
        <groupId>com.fasterxml.jackson.datatype</groupId>
        <artifactId>jackson-datatype-hibernate5</artifactId>
</dependency>

and finally add a new config

@Configuration
public class JacksonConfig {

@Bean
public Jackson2ObjectMapperBuilderCustomizer addCustomBigDecimalDeserialization() {
    return new Jackson2ObjectMapperBuilderCustomizer() {
        @Override
        public void customize(Jackson2ObjectMapperBuilder jacksonObjectMapperBuilder) {
            jacksonObjectMapperBuilder.featuresToDisable(DeserializationFeature.FAIL_ON_UNKNOWN_PROPERTIES);
            jacksonObjectMapperBuilder.modules(new Hibernate5Module());
        }

    };
}
}