问题描述
在 spring 中使用 hibernate 时，通常我们编码到如下界面
public interface UserRepository {
Collection<User> findByLastName(String lastName) throws DataAccessException;
//more methods
}
@Repository
public class JpaUserRepositoryImpl implements UserRepository {
@PersistenceContext
private EntityManager em;
@SuppressWarnings("unchecked")
public Collection<User> findByLastName(String lastName) {
Query query = this.em.createquery("SELECT disTINCT user FROM User user WHERE user.lastName LIKE :lastName");
query.setParameter("lastName",lastName + "%");
return query.getResultList();
}
}
@Entity
@Table(name = "users")
public class User {
..
}
我们如何在 Panache 中做同样的事情？
public interface UserRepository extends PanacheRepositoryBase<User,Integer> {
Collection<User> findByLastName(String lastName) ;
//more methods
}
@ApplicationScoped
public class JpaUserRepositoryImpl implements UserRepository {
@Inject
private EntityManager em;
@SuppressWarnings("unchecked")
public Collection<User> findByLastName(String lastName) {
Query query = this.em.createquery("SELECT disTINCT user FROM User user WHERE user.lastName LIKE :lastName");
query.setParameter("lastName",lastName + "%");
return query.getResultList();
}
}
它抛出
[2] Unsatisfied dependency for type javax.persistence.EntityManager and qualifiers [@Default]
- java member: org....repository.jpa.JpaUserRepositoryImpl#em
- declared on CLASS bean [types=[org....repository.jpa.JpaUserRepositoryImpl,org....repository.UserRepository,io.quarkus.hibernate.orm.panache.PanacheRepositoryBase<org.....model.User,java.lang.Integer>,java.lang.Object],qualifiers=[@Default,@Any],target=org.....repository.jpa.JpaUserRepositoryImpl]
[
解决方法
你可以让你的生活更轻松：
@ApplicationScoped
public class UserRepository implements PanacheRepository<User,Integer> {
public List<User> findByLastName(String lastName) {
return list("lastName",lastName);
}
}
@ApplicationScoped
public class UserService {
@Inject
private UserRepository userRepository;
@Transactional
public List<User> findByLastName(String lastName) {
return userRepository.findByLastName(lastName);
}
}
使用 Panache，您实际上不需要使用此存储库模式（尽管您可以）。我现在喜欢的工作方式是这样的：
@Entity
// Because you have defined your own ID (Integer),you should extend PanacheEntityBase. If you don't have a specific reason for defining a custom ID,I'd recommend to extend PanacheEntity,you'll get a Long ID for free.
public class User extends PanacheEntity {
public String lastName;
public static List<User> findByLastName(String lastName) {
return list("lastName",lastName);
}
}
@ApplicationScoped
public class UserService {
@Transactional
public List<User> findByLastName(String lastName) {
return User.findByLastName(lastName);
}
}
请注意，当您使用 PanacheEntity 模式时，您应该将所有实例字段定义为“public”。 Panache 包装了这些字段，因此您将保留所需的封装！ （所以如果你为这个字段定义了一个自定义的 setter，当有人这样做时，setter 将被调用：
user.lastName = "VanderLastName";
我不得不承认，我花了一段时间才习惯它。因此，特别是当您不想对您习惯的 (Spring) 进行太多更改时，您可能最好实施 Repository 模式。好消息：这完全是您自己的选择！