From Gossip@caterpillar

Hibernate Gossip: 一對一（唯一外鍵關聯）

現在考慮每一個User配給一間Room，形成一對一，user表格透過room_id作為外鍵參考至room：

一對一

在表格建立方面，使用 多對一 中的表格建立語句就可以了：
    create table room (
        id bigint not null auto_increment,
        address varchar(255),
        primary key (id)
    )

    create table user (
        id bigint not null auto_increment,
        name varchar(255),
        room_id bigint unique,
        primary key (id)
    )


物件方面，User的實例會參考至Room實例，而Room實例也參考至User實例：
  • User.java
package onlyfun.caterpillar;

public class User {
    private Long id;
    private String name;
    private Room room;
 
    public User() {}

    public Long getId() {
        return id;
    }

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

    public String getName() {
        return name;
    }

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

    public Room getRoom() {
        return room;
    }

    public void setRoom(Room room) {
        this.room = room;
    }
}

  • Room.java
package onlyfun.caterpillar; 

public class Room { 
    private Long id; 
    private String address;
    private User user;
 
    public Room() {}
 
    public Long getId() { 
        return id; 
    } 

    public void setId(Long id) { 
        this.id = id; 
    } 
 
    public String getAddress() { 
        return address; 
    }
 
    public void setAddress(String address) { 
        this.address = address; 
    }

    public User getUser() {
        return user;
    }

    public void setUser(User user) {
        this.user = user;
    }
}

使用外鍵來完成一對一，其實就是限制多對一關係中，「多」的一方只能有一個參考至「一」的一方，也就是多對一關係的一個特例，這可以在映射文件中使用 <many-to-one>標籤時，加上"unique"屬性來設定，例如：
  • User.hbm.xml
<?xml version="1.0" encoding="utf-8"?> 
<!DOCTYPE hibernate-mapping 
 PUBLIC "-//Hibernate/Hibernate Mapping DTD 3.0//EN" 
 "http://hibernate.sourceforge.net/hibernate-mapping-3.0.dtd"> 

<hibernate-mapping> 

    <class name="onlyfun.caterpillar.User" table="user"> 
        <id name="id" column="id"> 
            <generator class="native"/> 
        </id> 

        <property name="name" column="name"/> 
 
        <many-to-one name="room" 
                     column="room_id" 
                     class="onlyfun.caterpillar.Room"
                     cascade="all"
                     outer-join="true"
                     unique="true"/> 
    </class> 

</hibernate-mapping>

到這邊為止，單向一對一的映射已經完成，如果要再完成雙向一對一的關係，則可以在Room.hbm.xml中使用<one-to-one>標籤來定義：
  • Room.hbm.xml
<?xml version="1.0" encoding="utf-8"?> 
<!DOCTYPE hibernate-mapping 
 PUBLIC "-//Hibernate/Hibernate Mapping DTD 3.0//EN" 
 "http://hibernate.sourceforge.net/hibernate-mapping-3.0.dtd"> 

<hibernate-mapping> 

    <class name="onlyfun.caterpillar.Room" table="room"> 
        <id name="id" column="id"> 
            <generator class="native"/> 
        </id> 

        <property name="address" column="address" /> 
 
        <one-to-one name="user"
                    class="onlyfun.caterpillar.User"
                    property-ref="room"/>
     </class> 

</hibernate-mapping>

在<one-to-one>中，property-ref告訴Hibernate，查詢出user並將其參考至room。

一個儲存的例子如下：
User user1 = new User();
user1.setName("bush");
Room room1 = new Room();
room1.setAddress("NTU-M8-419");
user1.setRoom(room1);
       
User user2 = new User();
user2.setName("caterpillar");
Room room2 = new Room();
room2.setAddress("NTU-M8-418");
user2.setRoom(room2);
       
Session session = HibernateUti.getSessionFactory().openSession();
Transaction tx = session.beginTransaction();
 
session.save(user1);
session.save(user2);
       
tx.commit();
session.close();

在查詢Room時，User也會一載入，例如：
Session session = HibernateUtil.getSessionFactory().openSession();
Room room = (Room) session.load(Room.class, new Long(23));
System.out.println(room.getUser().getName());
session.close();

上面的查詢程式，Hibernate將使用以下的SQL：
Hibernate:
    select
        room0_.id as id1_1_,
        room0_.address as address1_1_,
        user1_.id as id0_0_,
        user1_.name as name0_0_,
        user1_.room_id as room3_0_0_
    from
        room room0_
    left outer join
        user user1_
            on room0_.id=user1_.room_id
    where
        room0_.id=?