【发布时间】：2014-12-17 21:39:42
【问题描述】：
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，我得到了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;
}
}
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。
【问题讨论】：