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具有@DiscriminatorValue的多个对象的单个存储库

[英]Single repository for multiple objects with @DiscriminatorValue

I have several entities sharing the same database table with @DiscriminatorValue with one base class: 我有几个实体使用一个基类与@DiscriminatorValue共享同一数据库表:

@Entity(name = "base_details")
@DiscriminatorColumn(name = "type")
@Inheritance(strategy = InheritanceType.SINGLE_TABLE)
@Getter
@Setter
public abstract class BaseDetails {

    @Id
    @Column(name = "identifier")
    private String identifier;

    @Column(name = "item_type")
    @Enumerated(EnumType.STRING)
    private Type type;

}

@Entity
@DiscriminatorValue("ONE")
@Getter
@Setter
public class FirstDetails extends BaseDetails {
...
}

@Entity
@DiscriminatorValue("TWO")
@Getter
@Setter
public class SecondDetails extends BaseDetails {
...
}

This works fine but I am running into an issue with having too many separate repositories: 这可以正常工作,但是我遇到了太多单独的存储库的问题:

@Repository
public interface OneDetailsRepository extends JpaRepository<OneDetails, String> { ... }

etc. 等等

Is there a way how to have a single repository (eq. BaseDetailsRepository ) which would work with all derived entities? 有没有一种方法可以拥有一个可用于所有派生实体的单一存储库(即BaseDetailsRepository )?

EDIT: What I am looking for is something like this: 编辑:我正在寻找的是这样的:

@Repository
public interface BaseDetailsRepository extends JpaRepository<? extends BaseDetails, String> { 
    OneDetails findOne(String identifier);
    SecondDetails findOne(String identifier);
}
@NoRepositoryBean    // Read only repository
public interface BaseDetailsRepository<T extends BaseDetails> 
extends JpaRepository<T, String> {  
 T findOne(String identifier);
 T findOne(Long id);
 Iterable<T> findAll();
 Iterable<T> findAll(Sort sort);
 Page<T> findAll(Pageable pageable);
}

and you can have also three repositories for each entity: 并且每个实体还可以具有三个存储库:

 @Transactional
public interface BaseRepository extends BaseDetailsRepository<BaseDetails> {   }
@Transactional
public interface FirstDetailsRepository extends BaseDetailsRepository<BaseDetails>,JpaRepository<T extends BaseDetails, String>{ }

@Transactional
public interface SecondDetailsRepository extends BaseDetailsRepository<BaseDetails>,JpaRepository<T extends BaseDetails, String> {  }

You can do as follows : 您可以执行以下操作:

Repository : BaseDetailsRepo.java 仓库:BaseDetailsRepo.java

@Repository
public interface BaseDetailsRepo extends JpaRepository<BaseDetails, String>{
//make the return according to you ..it shouldn't be list....

    @Query("FROM SecondDetails AS bd WHERE bd.identifier=:identifier")
    public List<SecondDetails> getSecondDetailsByIdentifier(@Param("identifier") String identifier);

    @Query("FROM FirstDetails AS bd WHERE bd.identifier=:identifier")
    public List<FirstDetails> getFirstDetailsByIdentifier(@Param("identifier") String identifier);          
}

Controller : DemoController.java 控制器:DemoController.java

@Controller
public class DemoController {

    @Autowired
    private BaseDetailsRepo baseDetailsRepo;


    @GetMapping(value="/test")
    public ResponseEntity test(){

        FirstDetails fd = new FirstDetails();
        fd.setIdentifier("fd1-demo");

        FirstDetails fd1 = new FirstDetails();
        fd1.setIdentifier("fd2-demo");


        SecondDetails sd = new SecondDetails();
        sd.setIdentifier("sd1-demo");
        baseDetailsRepo.save(fd);
        baseDetailsRepo.save(fd1);
        baseDetailsRepo.save(sd);

        return new ResponseEntity("Success",HttpStatus.OK);
    }


    @GetMapping(value="/test/second/{id}")
    public ResponseEntity getSecondDetails(@PathVariable String id){
        return new ResponseEntity(baseDetailsRepo.getSecondDetailsByIdentifier(id),HttpStatus.OK);
    }


    @GetMapping(value="/test/first/{id}")
    public ResponseEntity getFirstDetails(@PathVariable String id){
        return new ResponseEntity(baseDetailsRepo.getFirstDetailsByIdentifier(id),HttpStatus.OK);
    }
}

Output 输出量

测试

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