How to autowire SimpleJpaRepository in a spring boot application? - spring-boot

Question

I'm working on a project that uses SpringBoot 2.0.5 version, Spring Data JPA to persists and retrieve records using JPA. I autowired SimpleJpaRepository in the service layer. But while starting my application, it failed with
"NoSuchBeanDefinitionException"- No qualifying bean of type
'org.springframework.data.jpa.repository.support.SimpleJpaRepository<?, ?>'
available: expected at least 1 bean which qualifies as autowire candidate.
My controller, service and DAO are like below
Controller class:
#Controller
public class MyController{
#Autowired
private MyService<Person,PersonPK> service;
Service layer as
public interface MyService<V,K>{
methods defined
}
#Service("service")
public class MyServiceImpl<V,K> implements MyService<V,K>{
#Autowired
private SimpleJpaRepository<V,K> repository; // This dependency is failing
}
Application as :
#SpringBootApplication (exclude = {SecurityAutoConfiguration.class})
#EnableJpaRepositories
public class MyApplication {
public static void main(String[] args) {
SpringApplication.run(MyApplication.class, args);
}
}
Is my approach incorrect? Is that not the correct way of autowiring the SimpleJpaRepository.
There is no need for me to extend SimpleJpaRepository as Spring provided JPARepository is good for me, for now.
Thanks

Answer 1

You still need to create a repository interface that extends JpaRepisitory, or the spring repository type of your choice.
To quote the spring data documentation:
1.2.1 Defining repository interfaces
As a first step you define a domain class-specific repository interface. The interface must extend
Repository and be typed to the domain class and an ID type. If you
want to expose CRUD methods for that domain type, extend
CrudRepository instead of Repository.
Once you do create a new repository type, you will autowire by that type rather than the SimpleJpaRepository.

Answer 2

One way to get an implementation of the SimpleJpaRepository is by using a Configuration class to create an instance as a bean that you will use inside your service.
#Configuration
public class PersistanceConfiguration {
#PersistenceContext
private EntityManager entityManager;
#Bean
public SimpleJpaRepository<YourEntity, Long> getYourEntitySimpleRepository() {
return new SimpleJpaRepository<>(YourEntity.class, entityManager);
}
}
And inject it to your service as you would do with a JpaRepository, for example:
#Service
public class YourEntityServiceImpl<YourEntity, Long> implements YourEntityService {
private JpaRepository<YourEntity, K> repository;
private SimpleJpaRepository<YourEntity, K> simpleRepository;
#Autowired
public YourEntityServiceImpl(YourEntityRepository repository, SimpleJpaRepository<YourEntity, Long> simpleRepository) {
this.repository = repository;
this.simpleRepository = simpleRepository;
}
}

Answer 3

You should create a repository interface that extending JpaRepisitory.
#Repository
public interface MyRepository extends JpaRepisitory<T, ID> {
//
}
And the you should Autowired in your service class.
#Service("service")
public class MyServiceImpl<V,K> implements MyService<V,K>{
#Autowired
private MyRepository myRepository;
}