如何解决Spring Data JPA中的LazyInitializationException？

Question

我必须拥有一对多关系的类.当我尝试访问延迟加载的集合时,我得到了LazyInitializationException.我现在在网上搜索了一段时间,现在我知道我得到了异常,因为用于加载保存集合的类的会话已关闭.但是我没有找到解决方案(或者至少我不理解它们).基本上我有这些课程:

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;
    }

}

UserRepository

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

UserService

@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> {}

JobService

@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());
    }

}

我经常读到会话必须绑定到当前线程,但我不知道如何使用Spring的基于注释的配置来完成此操作.有人可以指出我该怎么做？

PS我想使用延迟加载,因此急切加载是没有选择的.

Answer 1

基本上,您需要在事务内部获取延迟数据.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了懒惰的集合,你将得到该异常,这在你的main()方法行中System.out.println(random.getCreatedJobs());.

现在,它归结为您的服务方法需要返回的内容.如果userService.getRandomUser()预期返回一个初始化了作业的用户,以便你可以操作它们,那么该方法就有责任获取它.使用Hibernate最简单的方法就是调用Hibernate.initialize(user.getCreatedJobs()).

Answer 2

考虑使用JPA 2.1和实体图：

延迟加载通常是 JPA 2.0 的一个问题。您必须在实体 FetchType.LAZY 或 FetchType.EAGER 上定义，并确保在事务中初始化关系。

这可以通过以下方式完成：

• 使用读取实体的特定查询
• 或者通过访问业务代码中的关系（每个关系的附加查询）。

这两种方法都远非完美，JPA 2.1 实体图是更好的解决方案：

• http://www.thoughts-on-java.org/jpa-21-entity-graph-part-1-named-entity/
• http://www.thoughts-on-java.org/jpa-21-entity-graph-part-2-define/

Answer 3

改变

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

到

@OneToMany(fetch = FetchType.EAGER, mappedBy = "creator")
private Set<Job> createdJobs = new HashSet<>();

或者在你的服务中使用 Hibernate.initialize ，它具有相同的效果。