大家都知道hibernate中的one-to-one映射主要有两种策略，（1）一对一主键关联（单向和双向）。（2）一对一外键映射（单项和双向）。本文主要讲解一下，一对一外键映射中的双向问题，在此前先通过一个实例了解。

  person和idCard,是一种一对一的关系，其中
 
t_person表                              
id       name       idCard(unique)
1         张三     
2         王五       1

其中王五是没有idcard，这也符合现实中的，有些人是没有身份证的。
t_idCard表
id         cardNo
1    11111111111111


实体类：
IdCard
package com.bjpowernode.hibernate;

public class IdCard {

private int id;

private String cardNo;

private Person person;

public Person getPerson() {
return person;
}

public void setPerson(Person person) {
this.person = person;
}

public int getId() {
return id;
}

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

public String getCardNo() {
return cardNo;
}

public void setCardNo(String cardNo) {
this.cardNo = cardNo;
}


}

Person
package com.bjpowernode.hibernate;

public class Person {

private int id;

private String name;

private IdCard idCard;

public int getId() {
return id;
}

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

public String getName() {
return name;
}

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

public IdCard getIdCard() {
return idCard;
}

public void setIdCard(IdCard idCard) {
this.idCard = idCard;
}
}

（3）配置文件
   IdCard的：
  <?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="com.bjpowernode.hibernate">
<class name="IdCard" table="t_idCard">
<id name="id">
<generator class="native"/>
</id>
<property name="cardNo"/>
<one-to-one name="person"  class="Person" />
</class>
</hibernate-mapping>

Person的：

<?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="com.bjpowernode.hibernate">
<class name="Person" table="t_person">
<id name="id">
<generator class="native"/>
</id>
<property name="name"/>
<many-to-one name="idCard" cascade="all"class="IdCard" unique="true" column="card_ID" />
</class>
</hibernate-mapping>
（3）
向t_person中插入数据：
import org.hibernate.Session;

import junit.framework.TestCase;

public class One2OneTest extends TestCase {

public void testSave1() {
Session session = null;
try {
session = HibernateUtils.getSession();
session.beginTransaction();

Person person = new Person();
person.setName("张三");

session.save(person);
session.getTransaction().commit();
}catch(Exception e) {
e.printStackTrace();
session.getTransaction().rollback();
}finally {
HibernateUtils.closeSession(session);
}
}
结果如下：
mysql> select * from t_person;
+----+------+---------+
| id | name | card_ID |
+----+------+---------+
|  1 | 张三 |    NULL |
+----+------+---------+
1 row in set (0.00 sec)

mysql> select * from t_idcard;
Empty set (0.00 sec)

在插入数据：
public void testSave2() {
Session session = null;
try {
session = HibernateUtils.getSession();
session.beginTransaction();

IdCard idCard = new IdCard();
idCard.setCardNo("1111111111");
session.save(idCard);

Person person = new Person();
person.setName("王五");
//建立关联
person.setIdCard(idCard);

session.save(person);

session.getTransaction().commit();
}catch(Exception e) {
e.printStackTrace();
session.getTransaction().rollback();
}finally {
HibernateUtils.closeSession(session);
}
}
数据库中的结果如下：
mysql> select * from t_person;
+----+------+---------+
| id | name | card_ID |
+----+------+---------+
|  1 | 张三 |    NULL |
|  2 | 王五 |       1 |
+----+------+---------+
2 rows in set (0.00 sec)

mysql> select * from t_idcard;
+----+------------+
| id | cardNo     |
+----+------------+
|  1 | 1111111111 |
+----+------------+
1 row in set (0.00 sec)
（4）加载数据，这样的话就可以从person的一端加载到idCard，
如下：
public void testLoad1() {
Session session = null;
try {
session = HibernateUtils.getSession();
session.beginTransaction();
Person person = (Person)session.load(Person.class, 2);
System.out.println("person.name=" + person.getName());
System.out.println("person.cardNo="+                  person.getIdCard().getCardNo());
session.getTransaction().commit();
}catch(Exception e) {
e.printStackTrace();
session.getTransaction().rollback();
}finally {
HibernateUtils.closeSession(session);
}
}
结果如下：
    Hibernate: select person0_.id as id0_0_, person0_.name asname0_0_, person0_.card_ID as card3_0_0_ from t_person person0_ whereperson0_.id=?
person.name=王五
        Hibernate: select idcard0_.id asid1_1_, idcard0_.cardNo as cardNo1_1_, person1_.id as id0_0_, person1_.name asname0_0_, person1_.card_ID as card3_0_0_ from t_idCard idcard0_ left outer joint_person person1_ on idcard0_.id=person1_.id where idcard0_.id=?
    person.cardNo=1111111111 这样的就找到了person对应的idcard，那能不能有idCard找到person呢？

public void testLoad2() {
Session session = null;
try {
session = HibernateUtils.getSession();
session.beginTransaction();
//                     
IdCard idCard = (IdCard)session.load(IdCard.class, 1);
System.out.println("idCard.cardNo=" + idCard.getCardNo());
System.out.println("idCard.person.name=" +idCard.getPerson().getName());
session.getTransaction().commit();
}catch(Exception e) {
e.printStackTrace();
session.getTransaction().rollback();
}finally {
HibernateUtils.closeSession(session);
}
}
结果如下：
Hibernate: select idcard0_.id as id1_1_, idcard0_.cardNo as cardNo1_1_,person1_.id as id0_0_, person1_.name as name0_0_, person1_.card_ID ascard3_0_0_ from t_idCard idcard0_ left outer join t_person person1_ onidcard0_.id=person1_.id where idcard0_.id=?
idCard.cardNo=1111111111
idCard.person.name=张三
结果对吗？
  肯定不对，idCard.cardNo=1111111111这是王五的idCard，怎么查出来张三的呢？原因在于：

idCard的配置文件问题：
应该在idCard的配置文件的<one-to-one name="person" class="Person"  />
修改为<one-to-one name="person" class="Person"  property-ref="idCard"/>
因为：如果不修改则idCard会根据自己的id和person中的id比较（因为one-to-one是通过id查找到的），这样是不符合要求的，因为我们t_idcard中的id和t_person中 Card_ID相比较，这样的话可以通过
property-ref="idCard" 的设置找到t_person表中Card_ID和它作比较找到我们要找的数据。