06_多对一关联映射[转]

  多对一的例子 , 多个员工 对 一个部门

package dao.po;   
/**  
 * @author zl 员工类  
 */  
public class Employee   
{   
 private int   id; // 员工的ID    
 private String  name; // 员工的名称   
 private Department depart;    //员工所在部门, 是多对一关系  

package dao.po;   
/**  
 
 * 部门类  
 *   
 * @author zl  
 *   
 */  
public class Department   
{   
    private int     id; //部门的ID    
    private String  name;   //部门的名称   
}

  

映射文件 :

<?xml version="1.0"?>  
<!DOCTYPE hibernate-mapping PUBLIC    
    "-//Hibernate/Hibernate Mapping DTD 3.0//EN"   
    "http://hibernate.sourceforge.net/hibernate-mapping-3.0.dtd">  
<hibernate-mapping    
    package="dao.po">  
    <class name="Employee">      
        <id name="id">  
            <generator class="native"/>  
        </id>  
        <property name="name" not-null="true" length="255" column="`name`"/>  
        <!-- 这里做多对一映射    -->  
        <!-- name="depart"是Employee类的属性名 -->  
        <!-- column="depart_id" 是表中字段名 -->  
        <!-- 注意:下面没有非空 约束 , 很多情况,外键是有非空约束的, 一会后面说-->  
        <!--column="depart_id" 中的depart_id是Employee对象的depart属性映射为Employee表中的depart_id字段-->  
        <many-to-one name="depart" column="depart_id" ></many-to-one>  
    </class>  
</hibernate-mapping> 

 

<?xml version="1.0"?>  
<!DOCTYPE hibernate-mapping PUBLIC    
    "-//Hibernate/Hibernate Mapping DTD 3.0//EN"   
    "http://hibernate.sourceforge.net/hibernate-mapping-3.0.dtd">  
<hibernate-mapping    
    package="dao.po">  
    <class name="Department">    
        <id name="id">  
            <generator class="native"/>  
        </id>        
        <property name="name" not-null="true" length="255" column="`name`"/>  
  
</class>  
</hibernate-mapping>  

保存员工 和 部门: 

 注意:保存的顺序: 

 

第一种: 

            session.save(depart); //先插入部门
            session.save(em1); //后插入员工, 因为员工的外键是 部门
            session.save(em2);

输出的SQL:

            Hibernate: insert into Department (`name`) values (?)
            Hibernate: insert into Employee (`name`, depart_id) values (?, ?)
            Hibernate: insert into Employee (`name`, depart_id) values (?, ?)


第二种: 

            session.save(em1); //先插入员工
            session.save(em2);
            session.save(depart); //后插入部门

输出的SQL:

            Hibernate: insert into Employee (`name`, depart_id) values (?, ?)
            Hibernate: insert into Employee (`name`, depart_id) values (?, ?)
            Hibernate: insert into Department (`name`) values (?)
            Hibernate: update Employee set `name`=?, depart_id=? where id=?
            Hibernate: update Employee set `name`=?, depart_id=? where id=?

 

            这里多了两句 update,要 注意.

 

如果Employee.hbm.xml 中外键有非空约束 ,如下:保存时只能用"第一种 "顺序.  用了第二种,先插入员工,但depart_id字段为空,会异常,不能插入.

<many-to-one name="depart" column="depart_id" not-null="true"></many-to-one>  

查询一个员工, 注意"部门": 

/**  
     * 测试查询  
     *   
     * @param args  
     */  
    public static void main(final String[] args)   
    {   
        //add(); //添加两个员工, 一个部门   
        final Employee em = query(7);   
        //员工的部门 , 执行Hibernate.initialize()后, 在session关闭前就取得了部门.   
        //若没有执行Hibernate.initialize(), 下面会抛异常.   
        System.out.println(em.getDepart().getName());   
}   
  
    /**  
     * 查询一个员工出来  
           */  
    public static Employee query(final int id)   
    {   
        Session session = null;   
        try  
        {   
            session = HibernateUtil.getSeesion();   
            final Transaction tx = session.beginTransaction();   
            final Employee em = (Employee) session.get(Employee.class, id); //按ID查   
            //因为 员工的 "部门" 属性会懒加载,   
            //在session关闭后,调用em.getDepart()无法取到部门信息   
            //所以这里用 Hibernate.initialize(em.getDepart()) 提前加载一下.   
            //是em.getDepart() 而不是em.   
            Hibernate.initialize(em.getDepart());   
            tx.commit();   
        return em;   
        }   
    finally  
        {   if (session != null)   
            {   
            session.close();   
            }   
    }   
  
}  

输出的SQL是:

Hibernate: select employee0_.id as id1_0_, employee0_.`name` as name2_1_0_, employee0_.depart_id as depart3_1_0_ from Employee employee0_ where employee0_.id=?

 

Hibernate: select department0_.id as id0_0_, department0_.`name` as name2_0_0_ from Department department0_ where department0_.id=?

 

做了两次查询, 没有使用 join on ,  想使用join on 还要配置. 默认不用.

 

关于懒加载,  会在以后的文章中专门讲, 这里先不讲, 只是用了用 Hibernate.initialize(em.getDepart()) 提前加载