java - 在 JpaRepository 中，如果一个参数对于许多查询方法来说是通用的，那么如何在 Repository 和 Service 中尽可能地做到最好

Question

public interface UserRepo extends JpaRepository<User, Long> {

    public List<User> findById(long id);

    public List<User> findByEmail(String email);

    public List<User> findByEmailAndCode(String email, Code code);

    public List<User> findByEmailAndClassType(String email, ClassType code);

}

public class UserService {
    @Autowired
    UserRepo userRepo;

    public List<user> fetchByClassType(ClassType ct) {
        return userRepo.findByEmailAndClassType("email", ct);
    }

}

here email need to fetch once how to avoid it many time to go on database or any other solution in controller it need to give again and again in each request mapping ...suggest

Answer 1

您可以选择规范，而不是创建一堆类似的存储库方法。

请参阅此处的文档:https://docs.spring.io/spring-data/jpa/docs/current/reference/html/#specifications

您必须扩展 JpaSecificationExecutor:

public interface CustomerRepository extends CrudRepository<Customer, Long>, JpaSpecificationExecutor {
    ...
}

然后您将获得一个接受规范的 findAll() 方法。

创建如下规范:

 public static Specification<Customer> isLongTermCustomer() {
    return new Specification<Customer>() {
      public Predicate toPredicate(Root<Customer> root, CriteriaQuery<?> query,
            CriteriaBuilder builder) {

         LocalDate date = new LocalDate().minusYears(2);
         return builder.lessThan(root.get(_Customer.createdAt), date);
      }
    };
  }

或者，如果您不喜欢 JPA Criteria API，您也可以使用 QueryDSL:

https://docs.spring.io/spring-data/jpa/docs/current/reference/html/#core.extensions.querydsl