Question

我刚刚开始了解FetchType Lazy和Eager。我理解差异，但即使我设置为Lazy，它仍然试图获取相关数据。

关系：1人：许多电话

查看研究尝试和教程：

https://www.mkyong.com/hibernate/hibernate-one-to-many-relationship-example-annotation/ http://howtodoinjava.com/hibernate/lazy-loading-in-hibernate/ https://howtoprogramwithjava.com/hibernate-eager-vs-lazy-fetch-type/

我理解得到相关数据，我仍然需要在会话中做到这一点（）所以对于我的特殊例子在我的Dao中，我需要这样的东西

  List<Person> persons = criteria.list();

    for(Person person : persons){

        Set sets = person.getPhones();

    }
    return persons;

到目前为止正确吗？

但问题是我没有在Dao，controller ..等任何地方调用person.getPhones（）但我得到了LazyInitializationEception。对于我的生活似乎无法捕捉到什么是错的。

堆栈跟踪

Jun 19, 2017 2:24:01 PM org.apache.catalina.core.StandardWrapperValve invoke
 SEVERE: Servlet.service() for servlet [app-dispatcher] in context with path 
 [/uni-starter-onetomany] threw exception [Request processing failed; nested exception is org.hibernate.LazyInitializationException: failed to lazily initialize a collection of role: com.app.person.Person.phones, could not initialize proxy - no Session] with root cause
org.hibernate.LazyInitializationException: failed to lazily initialize a collection of role: com.app.person.Person.phones, could not initialize proxy - no Session

Person.class

@Entity
@Table(name="person")
@Component
public class Person {

@Id
@GeneratedValue
private int person_id;

private String name;

private String age;

@OneToMany(cascade = CascadeType.ALL, fetch = FetchType.LAZY, mappedBy = "person")
private Set<Phone> phones;

//  Getter and setters

// ToString method by field name not by method

Phone.class

@Entity
@Table(name="phone")
@Component
public class Phone {

@Id
@GeneratedValue
private int phone_id;

private String type;

private String phone;

@ManyToOne(fetch = FetchType.LAZY)
@JoinColumn(name="person_person_id")
private Person person;

//  Getter and setters

PersonDao的

    @Repository
 @Transactional
 @Component("personDao")
 public class PersonDao {

@Autowired
private SessionFactory sessionFactory;

public Session session(){
    return sessionFactory.getCurrentSession();
}

public List<Person> getPersonList(){
    Criteria criteria = session().createCriteria(Person.class);

    List<Person> persons = criteria.list();

    //  for(Person person : persons){
    //
    //    Set sets = person.getPhones();
    //
    //  }

    return persons;
}


public void saveOrUpdate(Person person){
    session().saveOrUpdate(person);
}

}

控制器

    // Get list
@RequestMapping(path="/list", method = RequestMethod.GET, produces="application/json")
@ResponseBody
public Map<String, Object> getListPerson(){

    Map<String, Object> data = new HashMap<String, Object>();

    List<Person> persons = personDao.getPersonList();       

    data.put("results", persons);

    System.out.println(persons);

    return data;
}

问题

只要我不在会话中调用person.getPhones（），它应该只返回人吗？如果是这样的话，我可能会遇到LazyInitializationException的问题？
我还在Dao类中看过人们setFetchMode（）FetchMode.JOIN例如 Hibernate Criteria Join with 3 Tables 也许这可能是主观的但是什么是更好的做法？任何性能问题？

非常感谢任何想法，链接或文章...

更新感谢Abassa从Person.class中的toString中删除Phone in来解决问题。但我只是意识到，由于杰克逊，在序列化期间，它试图获取电话ojbect ....是否有解决方法？

Failed to write HTTP message: org.springframework.http.converter.HttpMessageNotWritableException: Could not write content: failed to lazily initialize a collection of role: com.app.person.Person.phones, could not initialize proxy - no Session

Answer 1

从Person类中的 toString 方法中删除phone字段。

致电时：

System.out.println(persons);

您尝试访问手机字段，因为println为列表中的每个人调用toString方法，因此您获得了LazyInitializationException。

Spring Hibernate FetchType LazyInitializationException即使不调用关联

1 个答案: