无法使MappedSuperClass抽象实体与Spring Boot和JPA一起使用

Question

I've done some research and I've gone through this guide .

Here is my structure of entities:

@MappedSuperClass
public abstract class Item{
    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    private long id;

    @Column(nullable = false)
    private String name;

    @OneToMany(cascade = CascadeType.ALL)
    @JoinColumn(name = "item_id")
    private Set<Picture> pictures = new HashSet<>();

}

Sample Item 1:

@Entity
public class Coke extends Item{

    @Column
    private String cokeProperty;

    @Column
    private String cokeProperty2;
}

Sample Item 2:

@Entity
public class Sprite extends Item{

    @Column
    private String spriteProperty;

    @Column
    private String spriteProperty2;
}

Picture class used by abstract class Item:

@Entity
public class Picture {

    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    private long id;

    private String name;
}

All of these have constructors, getters and setters, and toString overrides. I'm doing this because I want a table for Items , and then separate tables for Coke and Sprite , each containing a foreign key id from Items_table . Additionally, I assume that it would also create a table for Pictures that also has a foreign key from Items_table . What I can't figure out, however, is how to implement Spring Boot and JPA for this. I thought it would be as simple as this

My Repository for items which has a configured h2 database in application.properties

@Repository
public interface ItemRepository<Item, Long>{
}

My Service which implements ItemService interface and takes in an ItemRepository bean:

@Service
public class ItemServiceImpl implements ItemService{

private ItemRepository repository;

public ItemServiceImpl(ItemRepository repository) {
    this.repository = repository;
}

@Override
public ResponseEntity<List<Item>> getAll(){
    return new ResponseEntity<>(repository.findAll(), HttpStatus.OK);
}

@Override
public ResponseEntity<Item> getById(Long id){
    Optional<Item> item = repository.findById(id);
    return new ResponseEntity<>(item.orElse(null), HttpStatus.OK);
}

@Override
public ResponseEntity<Item> create(Item item){
    return new ResponseEntity<>(repository.save(item), HttpStatus.CREATED);
}    }

And then my Controller which takes in an ItemService bean:

@RestController
@RequestMapping("/api/items")
public class ItemController{

    private ItemService service;

    public ItemController(ItemService service) {
        this.service = service;
    }

    @GetMapping("")
    public ResponseEntity<List<Item>> findAll(){
        return service.getAll();
    }

    @GetMapping("/{id}")
    public ResponseEntity<Item> findById(@PathVariable(name = "id") Long id){
        return service.getById(id);
    }

    @PostMapping("")
    public ResponseEntity<Item> save(@RequestBody Item item){
        return service.create(item);
    }
}

But when I POST with Json Data like

{
    "name": "Toyota Civic 3 days sale!!",
    "pictures": [
      {
        "name": "picture 1"
      },
      {
        "name": "picture 2"
      }
    ],
    "cokeProperty": "sweet",
    "cokeProperty2": "not healthy"
}

I get this error:

com.fasterxml.jackson.databind.exc.InvalidDefinitionException: Cannot construct instance of com.example.demo.entities.product.Item (no Creators, like default construct, exist): abstract types either need to be mapped to concrete types, have custom deserializer, or contain additional type information at [Source: (PushbackInputStream); line: 1, column: 1]

I've also tried another strategy on my abstract class:

@Entity
@Inheritance(strategy = InheritanceType.TABLE_PER_CLASS)

Still no cigar. Same error. Can someone enlighten me on the proper way of using Spring Boot and JPA for abstract entities that are extended by other entities?

Answer 1

javax.persistence's @MappedSuperClass is not directly related with this issue. As the error message points it out, the cause is in jackson-databind.

Since Item is an abstract class, Jackson cannot determine which concrete class it should instantiate in order to deserialize the entity from JSON. That is why Item must be annotated with references to the implementations:

import com.fasterxml.jackson.annotation.JsonSubTypes;
import com.fasterxml.jackson.annotation.JsonTypeInfo;

@JsonTypeInfo(
  use = JsonTypeInfo.Id.NAME,
  include = JsonTypeInfo.As.PROPERTY, property = "type"
)
@JsonSubTypes({
    @JsonSubTypes.Type(value = Coke.class, name = "coke"),
    @JsonSubTypes.Type(value = Sprite.class, name = "sprite")
})
public abstract class Item {

   // no changes

}

A small change will also be necessary in the JSON, because it should point to the implementing type:

{
  "name": /* no changes  */,
  "type": "coke", /* new property */
  "pictures": /* no changes  */
}

Please note that @JsonTypeInfo provides a few choices for how to bind it with the implementation, not necessarily with the "type" field. There are various examples and documentation available about it.

A short test allows shows that Item is now deserialized properly:

String json = "{\n" +
    "    \"name\": \"Toyota Civic 3 days sale!!\",\n" +
    "    \"type\": \"coke\"," +
    "    \"pictures\": [\n" +
    "      {\n" +
    "        \"name\": \"picture 1\"\n" +
    "      },\n" +
    "      {\n" +
    "        \"name\": \"picture 2\"\n" +
    "      }\n" +
    "    ],\n" +
    "    \"cokeProperty\": \"sweet\",\n" +
    "    \"cokeProperty2\": \"not healthy\"\n" +
    "}";

ObjectMapper mapper = new ObjectMapper();

Item item = mapper.readValue(json, Item.class);

System.out.println(item);

// outputs:
// Coke(super=Item(id=0, name=Toyota Civic 3 days sale!!, pictures=[Picture(id=0, name=picture 2), Picture(id=0, name=picture 1)]), cokeProperty=sweet, cokeProperty2=not healthy)

But this change would not be enough to make things work for your service. As @Rujal Shrestha already mentions in his comment, there are also issues with repositories definitions and missing autowiring.