假设我有两个类:Subject和Client,Subject是基类。
@Entity
public class Client extends Subject
现在我想添加自定义的Jpa基本接口,因此可以在子接口中访问方法:
@NoRepositoryBean
public interface SubjectRepository <T extends Subject> extends
JpaRepository<T, Long>, CustomSubjectRepository<T> {}
CustomSubjectRepository看起来像:
public interface CustomSubjectRepository<T extends Subject> {
void saveEncrypted(T subject);
}
我需要实现所以我声明类:
@Repository
@Transactional
public class CustomSubjectRepositoryImpl<T extends Subject> implements
CustomSubjectRepository<T> {
@PersistenceContext
private EntityManager entityManager;
@Override
public void saveEncrypted(T subject) {
//implementation
}
}
然后想要创建ClientRepository并从SubjectRepository继承以访问saveEncrypted方法。
@Repository
public interface ClientRepository extends SubjectRepository<Client> {
}
但是当涉及到编译我得到:
创建名为'clientRepository'的bean时出错:init方法的调用失败;嵌套异常是java.lang.IllegalArgumentException:无法为方法public abstract void com.path.repositories.CustomSubjectRepository.saveEncrypted(com.path.models.Subject)创建查询!找不到类型Client的属性saveEncrypted!
您正在扩展接口,这样Spring将尝试创建名为saveEncrypted
的查询,而不是使用自定义方法。
我相信最好的解决方案是扩展班级CustomSubjectRepositoryImpl
。
@Repository
public class ClientRepository extends CustomSubjectRepositoryImpl<Client> {
}