Inferred type 'S' for type parameter 'S' is not within its bound; should extend 'ua.com.store.entity.Country

Question

I have a problem with my CountryServiceImpl,when I want realize method findOne in CountryServiceImpl it tells me "Inferred type 'S' for type parameter 'S' is not within its bound; should extend 'ua.com.store.entity.Country".

I wanted to fix by myself, but I don't understand what this means. Could you please help me with this issue.

Thank you.

@Entity
@Getter
@Setter
@NoArgsConstructor
@ToString
public class Country {

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

    private String countryName;

    @OneToMany(mappedBy = "country")
    private Set<Brand> brands = new HashSet<Brand>();
}


public interface CountryDAO extends JpaRepository<Country, Integer> {

    @Query("from Country c where c.countryName=:name")
    Country findByCountryName(@Param("name") String name);
}


public interface CountryService {

    void save(Country country);
    void delete(Country country);
    List<Country> findAll();
    Country findOne(int id);
    Country findByCountryName(String name);
}


@Service
public class CountryServiceImpl implements CountryService {

    @Autowired
    private CountryDAO dao;

    @Override
    public void save(Country country) {
        dao.save(country);
    }

    @Override
    public void delete(Country country) {
        dao.delete(country);
    }

    @Override
    public List<Country> findAll() {
        return dao.findAll();
    }

    @Override
    public Country findOne(int id) {
        return dao.findOne(id);
    }

    @Override
    public Country findByCountryName(String name) {
        return dao.findByCountryName(name);
    }
}

Capitulate

Answer 1

Spring documentation defines methods getOne as follows

<S extends T> Optional<S> findOne(Example<S> example)

In your method your input parameter is 'id' of type int but not bounded to interface Example.

To find an entity with it 'id' you can use the method

Optional<T> findById(ID id)

According to your implementation you may write it

@Override
public Country findOne(int id) {
    return dao.findById(id);
}

Lactometer

Answer 2

A 100% working solution is following:

@Override
public Country findOne(int id) {
    return dao.findById(id).orElse(null);
}

Repeater

Answer 3

It is possible to be relevant about spring-boot version. I meet the same issue when my spring-boot version is 2.0.1.RELEASE. But after change the spring-boot version to the 1.5.9.RELEASE, it is resolved.

Pingpingpong

Do you have some evidence to confirm this? – Wastrel

It is exists in the jar lib whose name is spring-data-commons. spring-data-commons jar lib is dependent to spring-boot-starter-data-jpa lib. When my dependencies is org.springframework.boot:spring-boot-starter-data-jpa:1.5.9.RELEASE, the code class org.springframework.data.repository.CrudRepository public interface CrudRepository<T, ID extends Serializable> extends Repository<T, ID> – Pingpingpong

When my dependencies is org.springframework.boot:spring-boot-starter-data-jpa:2.0.1.RELEASE, the code org.springframework.data.repository.query.QueryByExampleExecutor public interface QueryByExampleExecutor<T> And when 2.0.1.RELEASE version, there is no CrudRepository interface file. by the way, there is one sample at github.com/spring-guides/tut-rest/issues/53 . – Pingpingpong

Thanks for the replies, please edit them into your answer so they can help future visitors. :) – Wastrel

Answer 4

I just solved this problem in my program. I don't use the findOne method, I was just saving initial data with the save method. It turns out that I had copied my repo interface from another repo interface, and forgot to update the object in "extends JpaRepository<Object, Long>" to the new object.

Ojeda

thanks. without your answer, I may not think about this. – Asinine

Answer 5

You need to change from

public T getOne(ID id) {
        return repository.getOne(id);
}

To

public Optional<T> getOne(ID id) {
        return repository.findById(id);
}

Earthbound

getOne is method of JpaRepository and findById is method of CrudRepository, I think you will understand now – Earthbound

Answer 6

This Worked for me...

---

  @Override
    public Country findOne(int id) {
        return dao.findById(id).orElse(null);
    }

---

Graphite

Answer 7

I got the same problem.

ua.com.store.entity.Country -> It's like adding an external entity

You must import the correct directory of the entity "Counrty" that you will use in your project. e.g

import com.RomanSyhock.Entity.Country;

Sniffy

Answer 8

Additionally inherit another interface for Repository, namely JpaSpecificationExecutor

public interface CountryDAO extends JpaRepository<Country, Integer>, JpaSpecificationExecutor<Country> {

    @Query("from Country c where c.countryName=:name")
    Country findByCountryName(@Param("name") String name);
}

Oliveolivegreen