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);
    }
}

score 16 · Accepted Answer · answered Oct 03 '18 at 19:51

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);
}

score 4 · Answer 2 · edited Feb 28 '23 at 02:32

4

A 100% working solution is following:

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

edited Feb 28 '23 at 02:32

Prabhu

5,296
4
37
45

answered Aug 21 '19 at 16:20

Ihsan Ullah Khan

102
8

score 3 · Answer 3 · answered Mar 19 '19 at 07:57

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.

answered Mar 19 '19 at 07:57

yang

41
3

Do you have some evidence to confirm this? – JJJ Mar 19 '19 at 08:16
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 extends Repository` – yang Mar 19 '19 at 14:52
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` And when 2.0.1.RELEASE version, there is no CrudRepository interface file. by the way, there is one sample at https://github.com/spring-guides/tut-rest/issues/53 . – yang Mar 19 '19 at 14:52
Thanks for the replies, please edit them into your answer so they can help future visitors. :) – JJJ Mar 19 '19 at 14:52

score 3 · Answer 4 · answered Jul 17 '20 at 21:23

3

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.

answered Jul 17 '20 at 21:23

Sean Tunell

31
1

thanks. without your answer, I may not think about this. – Ridwan Sep 17 '21 at 08:02

score 1 · Answer 5 · answered Nov 22 '19 at 06:25

1

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);
}

answered Nov 22 '19 at 06:25

Shahid Hussain Abbasi

2,508
16
10

getOne is method of JpaRepository and findById is method of CrudRepository, I think you will understand now – Shahid Hussain Abbasi Nov 22 '19 at 06:39

score 1 · Answer 6 · answered Aug 04 '20 at 08:38

1

This Worked for me...

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

answered Aug 04 '20 at 08:38

Niroshan Ratnayake

3,433
3
20
18

score 0 · Answer 7 · answered Nov 04 '21 at 14:46

0

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;

answered Nov 04 '21 at 14:46

Tugba Alparslan

145
6

score 0 · Answer 8 · answered May 27 '23 at 02:52

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);
}

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

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