Question

我试图使用Spring引导程序来使用Spring JPA存储库。我对春天来说是全新的，我只知道Spring到目前为止暴露给我的那么多冬眠。这不是很多。这就是我正在尝试的

@Configuration
@EnableJpaRepositories
@Import(RepositoryRestMvcConfiguration.class)
@EnableAutoConfiguration
public class Application {

    public static void main(String[] args) {

        ConfigurableApplicationContext context = SpringApplication.run(Application.class);
        UserRepo repo = context.getBean(UserRepo.class);
        seedDB(repo);
        for (User u: repo.findAll())
            System.out.println(u);
    }   
    private static void seedDB(UserRepo repo) {
        Doctor jones = repo.save(new Doctor("Fred", "Jones"));
        Doctor smith = repo.save(new Doctor("Rebbecca", "Smith"));
        Patient balboa = repo.save(new Patient("Rocky", "Balboa", new Date()));
        Patient locke = repo.save(new Patient("John", "Locke", new Date()));
        Patient linus = repo.save(new Patient("Ben", "Linus", new Date()));
        balboa.addDr2Ways(jones);//this adds jones to balboa's list of doctors and balboa to jones' list of patients
        locke.addDr2Ways(smith);
        locke.addDr2Ways(jones);
        linus.addDr2Ways(smith);
        repo.save(jones);
        repo.save(smith);
        repo.save(balboa);
        repo.save(locke);
        repo.save(linus);    
    }

在我尝试打印的循环中，我得到一个hibernate延迟初始化异常，因为没有会话。我做错了什么？

以下是对评论的回应：

医生和患者都继承用户如下：

@Entity
public abstract class User {

@Id @GeneratedValue private long ID;
private String firstName, lastName;

public User(){}
public User(String firstName, String lastName){
    this.firstName = firstName;
    this.lastName = lastName;
}

...

    @Entity
public class Doctor extends User {
    public Doctor(){}
    public Doctor(String firstName, String lastName) {
        super(firstName, lastName);
    }
    @ManyToMany
    private List<Patient> patients = new ArrayList<>();
    public String toString(){
        String result = "dr " + super.getFirstName() + " has the following patients\n";
        for (Patient p: patients)
            result += p.getFirstName() + "\n";
        return result;
    }
    public void addPatient1Way(Patient patient) {
        patients.add(patient);

    }
}

    @Entity
public class Patient extends User {
    public Patient(){}
    public Patient(String firstName, String lastName, Date birthday) {
        super(firstName, lastName);
        this.birthday = birthday;
    }
    private Date birthday;

    @ManyToMany(mappedBy = "patients")
    private List<Doctor> doctors= new ArrayList<>();
    public String toString(){
        return "patient " + super.getFirstName();
    }
    public void addDr2Ways(Doctor dr) {
        // TODO Auto-generated method stub
        doctors.add(dr);
        dr.addPatient1Way(this);
    }
}

如何防止延迟初始化异常

0 个答案: