Java 类型参数“S”的推断类型“S”不在其范围内；应该扩展 'ua.com.store.entity.Country
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Inferred type 'S' for type parameter 'S' is not within its bound; should extend 'ua.com.store.entity.Country
提问by Roman Sychok
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".
我的 CountryServiceImpl 有问题，当我想在 CountryServiceImpl 中实现 findOne 方法时，它告诉我“类型参数 'S' 的推断类型 'S' 不在其范围内；应该扩展 '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.
我想自己修复，但我不明白这是什么意思。你能帮我解决这个问题吗？
Thank you.
谢谢你。
@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);
}
}
采纳答案by Lakshi
Spring documentation defines methods getOne as follows
Spring 文档定义了 getOne 方法如下
<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.
在您的方法中，您的输入参数是 int 类型的“id”，但不受接口示例的限制。
To find an entity with it 'id' you can use the method
要查找带有“id”的实体，您可以使用该方法
Optional<T> findById(ID id)
According to your implementation you may write it
根据您的实现，您可以编写它
@Override
public Country findOne(int id) {
return dao.findById(id);
}
回答by yang
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.
可能与 spring-boot 版本相关。当我的 spring-boot 版本是 2.0.1.RELEASE 时，我遇到了同样的问题。但是把spring-boot版本改成1.5.9.RELEASE后就解决了。
回答by Ihsan Ullah Khan
A 100% working solution is following:
100% 工作解决方案如下：
@Override
public Country findOne(int id) {
return dao..findById(id).orElse(null);
}
回答by Shahid Hussain Abbasi
You need to change from
你需要从
public T getOne(ID id) {
return repository.getOne(id);
}
To
到
public Optional<T> getOne(ID id) {
return repository.findById(id);
}
