java - 如何解决 Spring Data JPA 中的 LazyInitializationException？

Question

我有两个具有一对多关系的类。当我尝试访问延迟加载的集合时，我得到 LazyInitializationException。我已经在网上搜索了一段时间，现在我知道我得到了异常，因为用于加载包含集合的类的 session 已关闭。但是，我没有找到解决方案(或者至少我不理解它们)。基本上我有这些类(class):

用户

@Entity
@Table(name = "user")
public class User {

    @Id
    @GeneratedValue
    @Column(name = "id")
    private long id;

    @OneToMany(mappedBy = "creator")
    private Set<Job> createdJobs = new HashSet<>();

    public long getId() {
        return id;
    }

    public void setId(final long id) {
        this.id = id;
    }

    public Set<Job> getCreatedJobs() {
        return createdJobs;
    }

    public void setCreatedJobs(final Set<Job> createdJobs) {
        this.createdJobs = createdJobs;
    }

}

用户存储库

public interface UserRepository extends JpaRepository<User, Long> {}

用户服务

@Service
@Transactional
public class UserService {

    @Autowired
    private UserRepository repository;

    boolean usersAvailable = false;

    public void addSomeUsers() {
        for (int i = 1; i < 101; i++) {
            final User user = new User();

            repository.save(user);
        }

        usersAvailable = true;
    }

    public User getRandomUser() {
        final Random rand = new Random();

        if (!usersAvailable) {
            addSomeUsers();
        }

        return repository.findOne(rand.nextInt(100) + 1L);
    }

    public List<User> getAllUsers() {
        return repository.findAll();
    }

}

工作

@Entity
@Table(name = "job")
@Inheritance
@DiscriminatorColumn(name = "job_type", discriminatorType = DiscriminatorType.STRING)
public abstract class Job {

    @Id
    @GeneratedValue
    @Column(name = "id")
    private long id;

    @ManyToOne
    @JoinColumn(name = "user_id", nullable = false)
    private User creator;

    public long getId() {
        return id;
    }

    public void setId(final long id) {
        this.id = id;
    }

    public User getCreator() {
        return creator;
    }

    public void setCreator(final User creator) {
        this.creator = creator;
    }

}

JobRepository

public interface JobRepository extends JpaRepository<Job, Long> {}

工作服务

@Service
@Transactional
public class JobService {

    @Autowired
    private JobRepository repository;

    public void addJob(final Job job) {
        repository.save(job);
    }

    public List<Job> getJobs() {
        return repository.findAll();
    }

    public void addJobsForUsers(final List<User> users) {
        final Random rand = new Random();

        for (final User user : users) {
            for (int i = 0; i < 20; i++) {
                switch (rand.nextInt(2)) {
                case 0:
                    addJob(new HelloWorldJob(user));
                    break;
                default:
                    addJob(new GoodbyeWorldJob(user));
                    break;
                }
            }
        }
    }

}

应用

@Configuration
@EnableAutoConfiguration
@ComponentScan
public class App {

    public static void main(final String[] args) {
        final ConfigurableApplicationContext context = SpringApplication.run(App.class);
        final UserService userService = context.getBean(UserService.class);
        final JobService jobService = context.getBean(JobService.class);

        userService.addSomeUsers();                                 // Generates some users and stores them in the db
        jobService.addJobsForUsers(userService.getAllUsers());      // Generates some jobs for the users

        final User random = userService.getRandomUser();            // Picks a random user

        System.out.println(random.getCreatedJobs());
    }

}

我经常读到 session 必须绑定(bind)到当前线程，但我不知道如何使用 Spring 的基于注释的配置来做到这一点。有人能指出我该怎么做吗？

附:我想使用延迟加载，因此无法使用急切加载。

Answer 1

基本上，您需要在事务中获取惰性数据。如果您的服务类是 @Transactional，那么当您在其中时，一切都应该没问题。一旦你离开服务类，如果你尝试 get 惰性集合，你会得到那个异常，它在你的 main() 方法中，行 System.out.println(random.getCreatedJobs());.

现在，归结为您的服务方法需要返回什么。如果 userService.getRandomUser() 预计会返回一个已初始化作业的用户，以便您可以操作它们，那么获取它是该方法的责任。使用 Hibernate 最简单的方法是调用 Hibernate.initialize(user.getCreatedJobs())。