How to use ModelMapper for deep inheritance objects?

Question

A.java

@Entity
@Getter
@Setter
@Inheritance
@JsonTypeInfo(use=JsonTypeInfo.Id.NAME, inclue=JsonTypeInfo.As.PROPERTY, property="type") 
@JsonSubTypes({
    @JsonSubTypes.Type(value=AA.class,name="aa"),
    @JsonSubTypes.Type(value=AB.class,name="ab"),
})
public abstract class A {
    @Id
    @GeneratedValue
    private Long id;

    //json ignore for getter
    @ManyToOne
    private A parent;

    @OneToMany(mappedBy="parent")
    private List<A> children;
}

AA.java

@Entity
@Getter
@Setter
@DiscriminatorValue("aa")
public class AA extends A{
    private User user;
}

AB.java

@Entity
@Getter
@Setter
@DiscriminatorValue("ab")
public class AB extends A {
    private String name;
}

Now, when I return an instance of class AB as JSON, it looks like this:

{
     "id": 1,
     "type": "ab",
     "children": [...],
     "name": "ali"
}

Since I want to use a custom User JSON, I want to use ModelMapper. All things are Ok except Absence of type.

My DTOs:

@Getter
@Setter
public class ADto {
     private Long id;

     private List<ADto> children;
}

@Getter
@Setter
public class AADto extends ADto {
     private UserDto user;
}

@Getter
@Setter
public class ABDto extends ADto {
     private String name;
}

How can I use ModelMapper and get type of class?

Andros

Are you sure the example is correct? For example, AB is not used anywhere and it has only one filed but it's json has 4 fields. Could you reformat/shortly your questions so that it is clear what you need. – Glycerol

@OleksandrShpota you are right, AA and AB extends class A. – Andros

Answer 1

Possible solutions:

1. Have you tried to add the attribute type in the parent class? This should be enough if you retrieve this field from the DB.

2. A simple tested trick, add protected String type = this.getClass().getSimpleName().toLowerCase();. Notice that it must be protected. Here's my tested code:

   @Data
   abstract class A {
     protected String type = this.getClass().getSimpleName().toLowerCase();
   }
   
    class AB extends A {}
   
    class AA extends A {}
   
    @Test
    public void test(){
       A a1 = new AB();
       A a2 = new AA();
       assertThat(a1.getType()).isEqualTo("ab");
       assertThat(a2.getType()).isEqualTo("aa");
    }
   

Imray

Answer 2

You can InheritanceType when you define parent class which will help to show the child type of class

Ex- A.java

@Entity
@Getter
@Setter
@Inheritance(strategy=InheritanceType.JOINED) // change needed 
@JsonTypeInfo(use=JsonTypeInfo.Id.NAME, inclue=JsonTypeInfo.As.PROPERTY, property="type") 
@JsonSubTypes({
    @JsonSubTypes.Type(value=AA.class,name="aa"),
    @JsonSubTypes.Type(value=AB.class,name="ab"),
})
public abstract class A {

Ovotestis

Answer 3

There is now type map for solving this problem. See the type-map-inheritence documentation.

// Creates a base TypeMap with explicit mappings
TypeMap<BaseSrc, BaseDest> typeMap = modelMapper.createTypeMap(BaseSrc.class, BaseDest.class)
    .addMapping(BaseSrc::getFirstName, BaseDest::setName);

typeMap.include(SrcA.class, DestA.class)
    .include(SrcB.class, DestB.class)
    .include(SrcC.class, DestC.class);

Unawares