hibernate系列（四）一对一关联关系-白红宇

以Person类和IDCard类为例，这里仅仅说一种一对一关联关系，即Person类拥有IDCard，但是IDCard不含Person类，数据库库的表如下：

                         
CREATE
TABLE
`hibernate`.`person` (
  
`id` 
INT
NOT
NULL
AUTO_INCREMENT,
  
`
name
` 
VARCHAR
(45) 
NULL
,
  
`age` 
INT
NULL
,
  
`idcard_id` 
INT
NULL
,
  
PRIMARY
KEY
(`id`));
        
      

                         
CREATE
TABLE
`hibernate`.`idcard` (
  
`id` 
INT
NOT
NULL
AUTO_INCREMENT,
  
`number` 
INT
NULL
,
  
`content` 
VARCHAR
(45) 
NULL
,
  
PRIMARY
KEY
(`id`));
        
      

Person类如下：

                         
public
class
Person { 
    
private
Long id;
    
private
String name;
    
private
Long age;
    
private
IDCard idCard;
//省略get、set方法
}
        
      

Person类对应的Person.hbm.xml映射文件为：

                         
<
hibernate-mapping
>
    
<
class
name
=
"com.ligang.domain.Person"
table
=
"person"
>
        
<
id
name
=
"id"
column
=
"id"
type
=
"long"
>
            
<
generator
class
=
"identity"
/>
        
</
id
>
        
<
property
name
=
"name"
column
=
"name"
type
=
"string"
/>
        
<
property
name
=
"age"
column
=
"age"
type
=
"long"
/>
        
<
many-to-one
name
=
"idCard"
class
=
"com.ligang.domain.IDCard"
column
=
"idcard_id"
cascade
=
"save-update"
></
many-to-one
>
    
</
class
>
</
hibernate-mapping
>
        
      

虽然是一对一但是，这种形式的一对一就是多对一的特例，所以仍然使用<many-to-one>的标签，其中的name指的是Person类的idCard属性，column指的是person表中的字段名为idcard_id,class指的是将IDCard类对应的表的主键的值作为idcard_id的值。cascade字段表示保存Person类时级联的保存IDCard类。

下面看下IDCard类：

                         
public
class
IDCard { 
    
private
Long id;
    
private
Long number;
    
private
String content;
//省略get、set方法
}
        
      

IDCard类对应的IDCard.hbm.xml映射文件为：

                         
<
hibernate-mapping
>
    
<
class
name
=
"com.ligang.domain.IDCard"
table
=
"idcard"
>
        
<
id
name
=
"id"
column
=
"id"
type
=
"long"
>
            
<
generator
class
=
"identity"
/>
        
</
id
>
        
<
property
name
=
"number"
column
=
"number"
type
=
"long"
/>
        
<
property
name
=
"content"
column
=
"content"
type
=
"string"
/>
    
</
class
>
</
hibernate-mapping
>
        
      

然后就是测试增添方法：

                         
@Test
    
public
void
addPerson(){ 
        
Session session=hibernateDao.getSession();
        
Transaction tx=session.beginTransaction();
        
Person p=
new
Person();
        
p.setName(
"张三"
);
        
p.setAge(122L);
        
IDCard idCard=
new
IDCard();
        
idCard.setNumber(123445L);
        
idCard.setContent(
"你是一个人"
);
        
p.setIdCard(idCard);
        
session.save(p);
        
tx.commit();
        
session.close();
    
}

此时就会先保存IDCard对象，然后获取其主键并赋值给person表的idcard_id字段，然后增添Person对象。如下sql:

                         
Hibernate: 
insert
into
hibernate.idcard (number, content) 
values
(?, ?)
Hibernate: 
insert
into
hibernate.person (
name
, age, idcard_id) 
values
(?, ?, ?)
        
      

更新如下：

                         
@Test
    
public
void
updatePerson(){ 
        
Session session=hibernateDao.getSession();
        
Transaction tx=session.beginTransaction();
        
Person p=(Person) session.get(Person.
class
,6L);
        
p.setName(
"张三"
);
        
p.setAge(122L);
        
IDCard idCard=
new
IDCard();
        
idCard.setNumber(123445L);
        
idCard.setContent(
"你是一个人"
);
        
p.setIdCard(idCard);
        
session.save(p);
        
tx.commit();
        
session.close();
    
}

此时的sql如下：

                         
Hibernate: 
select
person0_.id 
as
id1_3_0_, person0_.
name
as
name2_3_0_, person0_.age 
as
age3_3_0_, person0_.idcard_id 
as
idcard_i4_3_0_ 
from
hibernate.person person0_ 
where
person0_.id=?
Hibernate: 
insert
into
hibernate.idcard (number, content) 
values
(?, ?)
Hibernate: 
update
hibernate.person 
set
name
=?, age=?, idcard_id=? 
where
id=?
        
      

此时并没有删除原有的IDCard，只是根据Person再也找不到它了。目前我还不知道怎么设置来删除无用的IDCard。

获取：由Person的主键获取IDCard比较容易，但是如果想从IDCard主键获取Person呢？

首先更改IDCard类，添加person属性：

                         
public
class
IDCard { 
    
private
Long id;
    
private
Long number;
    
private
String content;
    
private
Person person;
//略get、set方法
}
        
      

然后更改上述的IDCard.hbm.xml映射文件如下：

                         
<
hibernate-mapping
>
    
<
class
name
=
"com.ligang.domain.IDCard"
table
=
"idcard"
>
        
<
id
name
=
"id"
column
=
"id"
type
=
"long"
>
            
<
generator
class
=
"identity"
/>
        
</
id
>
        
<
property
name
=
"number"
column
=
"number"
type
=
"long"
/>
        
<
property
name
=
"content"
column
=
"content"
type
=
"string"
/>
        
<
one-to-one
name
=
"person"
property-ref
=
"idCard"
></
one-to-one
>
    
</
class
>
</
hibernate-mapping
>
        
      

添加了<one-to-one>标签，同时使用了property-ref属性，看下查询再解释:

                         
@Test
    
public
void
getPerson(){ 
        
Session session=hibernateDao.getSession();
        
Transaction tx=session.beginTransaction();
        
IDCard idCard=(IDCard) session.get(IDCard.
class
,2L);
        
System.out.println(idCard.getPerson().getName());
        
tx.commit();
        
session.close();
    
}

查询的sql如下：

                         
Hibernate: 
select
idcard0_.id 
as
id1_1_0_, idcard0_.number 
as
number2_1_0_, idcard0_.content 
as
content3_1_0_, person1_.id 
as
id1_3_1_, person1_.
name
as
name2_3_1_, person1_.age 
as
age3_3_1_, person1_.idcard_id 
as
idcard_i4_3_1_ 
from
hibernate.idcard idcard0_ 
left
outer
join
hibernate.person person1_ 
on
idcard0_.id=person1_.idcard_id 
where
idcard0_.id=?
张三
        
      

将Person查出来了，然后看下他是怎么查的，看下IDCard的<one-to-one name="person" property-ref="idCard">，name指的是IDCard的person属性，而property-ref指向了另一个类的属性，这里就是Person类的idCard属性，要想查出Person总要告诉hibernate idcard表的哪个字段和person表的哪个字段相连接吧，这里的property-ref="idCard"，即指定了要和Person类的idCard属性所对应的字段相连接，默认是采用主键来和该字段相连接的，我不知道能否指定。即idcard表的主键id和Person类的idCard属性对应的字段idcard_id相连接，来查询Person,看sql语句：hibernate.idcard idcard0_ left outer join hibernate.person person1_ on idcard0_.id=person1_.idcard_id

上述一对一的关联关系是通过外键来连接，他们也可以使用相同的主键实现一对一。这里不再讲述。