java - 如何避免 Hibernate 中的 LazyInitializationException？

Question

我将 Hibernate 用作具有多个外键关系的数据库的 ORM。问题是有时我想获取这些相关的数据集，有时我不想，所以在这些集合上我将“获取”设置为“惰性”。不幸的是，每次我尝试序列化这些对象时，Hibernate 都会抛出 LazyInitializationException，因为 session 已关闭。使用 OpenSessionInView 过滤器只会导致 Hibernate 无论如何都要填充这些集合，从而首先破坏了拥有惰性集合的全部目的。

有没有一种简单的方法可以序列化或提取 POJO 中填充的数据而不触发 LIE，并且不必填充所有惰性集合？

编辑:这是我试图开始工作的一些示例代码，处理两个表，“部门”和“员工”，这是与部门的一对多关系的 child 。我希望能够查看数据库中列出的部门，而不必加载属于所述部门的所有员工:

部门:

package com.test.model;
// Generated Apr 7, 2012 7:10:28 PM by Hibernate Tools 3.4.0.CR1

import java.util.HashSet;
import java.util.Set;
import javax.persistence.Column;
import javax.persistence.Entity;
import javax.persistence.FetchType;
import javax.persistence.GeneratedValue;
import static javax.persistence.GenerationType.IDENTITY;
import javax.persistence.Id;
import javax.persistence.OneToMany;
import javax.persistence.Table;

/**
 * Departments generated by hbm2java
 */
@Entity
@Table(name="Departments"
    ,catalog="test"
)
public class Departments  implements java.io.Serializable {


     private Integer id;
     private String name;
     private Set<Employees> employeeses = new HashSet(0);

    public Departments() {
    }


    public Departments(String name) {
        this.name = name;
    }
    public Departments(String name, Set employeeses) {
       this.name = name;
       this.employeeses = employeeses;
    }

     @Id @GeneratedValue(strategy=IDENTITY)


    @Column(name="Id", unique=true, nullable=false)
    public Integer getId() {
        return this.id;
    }

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


    @Column(name="Name", nullable=false)
    public String getName() {
        return this.name;
    }

    public void setName(String name) {
        this.name = name;
    }

    @OneToMany(fetch=FetchType.LAZY, mappedBy="departments")
    public Set<Employees> getEmployeeses() {
        return this.employeeses;
    }

    public void setEmployeeses(Set employeeses) {
        this.employeeses = employeeses;
    }
}

员工:

package com.test.model;
// Generated Apr 7, 2012 7:10:28 PM by Hibernate Tools 3.4.0.CR1


import javax.persistence.Column;
import javax.persistence.Entity;
import javax.persistence.FetchType;
import javax.persistence.GeneratedValue;
import static javax.persistence.GenerationType.IDENTITY;
import javax.persistence.Id;
import javax.persistence.JoinColumn;
import javax.persistence.ManyToOne;
import javax.persistence.Table;

/**
 * Employees generated by hbm2java
 */
@Entity
@Table(name="Employees"
    ,catalog="test"
)
public class Employees  implements java.io.Serializable {


     private Integer id;
     private Departments departments;
     private String firstName;
     private String lastName;

    public Employees() {
    }

    public Employees(Departments departments, String firstName, String lastName) {
       this.departments = departments;
       this.firstName = firstName;
       this.lastName = lastName;
    }

     @Id @GeneratedValue(strategy=IDENTITY)


    @Column(name="Id", unique=true, nullable=false)
    public Integer getId() {
        return this.id;
    }

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

    @ManyToOne(fetch=FetchType.LAZY)
    @JoinColumn(name="DepartmentsId", nullable=false)
    public Departments getDepartments() {
        return this.departments;
    }

    public void setDepartments(Departments departments) {
        this.departments = departments;
    }


    @Column(name="FirstName", nullable=false)
    public String getFirstName() {
        return this.firstName;
    }

    public void setFirstName(String firstName) {
        this.firstName = firstName;
    }


    @Column(name="LastName", nullable=false)
    public String getLastName() {
        return this.lastName;
    }

    public void setLastName(String lastName) {
        this.lastName = lastName;
    }
}

我的操作类(由 Struts2 XSLT 结果序列化):

package com.test.view;

import java.util.List;

import java.util.Iterator;

import com.opensymphony.xwork2.ActionSupport;
import com.test.controller.DepartmentsManager;
import com.test.model.Departments;
import com.test.util.HibernateUtil;

public class DepartmentsAction extends ActionSupport {
private DepartmentsManager departmentsManager;
private List<Departments> departmentsList;

public DepartmentsAction() {
    this.departmentsManager = new DepartmentsManager();
}

public String list() {
    this.departmentsList = departmentsManager.list();
    System.out.println("Execute called");
    HibernateUtil.createDTO(departmentsList);
    return SUCCESS;
}

public List<Departments> getDepartmentsList() {
    return departmentsList;
}

public void setDepartmentsList(List<Departments> departmentsList) {
    this.departmentsList = departmentsList;
}
}

我的经理类(Action 类调用它来填充部门列表):

package com.test.controller;

import java.util.List;

import java.util.Iterator;

import org.hibernate.Criteria;
import org.hibernate.Hibernate;
import org.hibernate.HibernateException;
import org.hibernate.Query;
import org.hibernate.Session;

import com.test.model.Departments;
import com.test.util.HibernateUtil;

public class DepartmentsManager {
public List<Departments> list() {
    Session session = HibernateUtil.getSessionFactory().getCurrentSession();
    session.beginTransaction();
    List<Departments> set = null;
    try {
        Query q = session.createQuery("FROM Departments");
        /*Query q = session.createQuery("FROM Departments d JOIN FETCH d.employeeses e");*/
        q.setResultTransformer(Criteria.DISTINCT_ROOT_ENTITY);
        set = (List<Departments>) q.list();
    } catch (HibernateException e) {
        e.printStackTrace();
        session.getTransaction().rollback();
    }
    session.getTransaction().commit();
    return set;
}
}

Answer 1

惰性集合仅在事务范围内工作(从数据库中检索拥有实体的范围)。换句话说，您不应该传递一个 Hibernate 实体，其中包含未加载的惰性子实体或事务范围之外的集合。

如果要将实体传递给 JSP、序列化代码或其他任何内容，则需要构建另一个实体或使用 lazy="false"。