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例如:表示主机HOST含有多个磁盘DISK关系:

在PD的OOM中双击一条关系连线,设置Navigate,这样才会在各自hbm.xml中生成many-to-one及one-to-many关系。



另外,根据“附文”的效率说明还需要设置:由多方DISK维护关系,并且当HOST更新时,也要更新DISK。



生成的HBM.XML如下:
HOST的:
      <joined-subclass name="Host" table="host" dynamic-update="false" dynamic-insert="false" select-before-update="false" lazy="true" abstract="false"> 
       <key on-delete="noaction" unique="true">
         <column name="dev_id" sql-type="int" not-null="true" length="0"/>
       </key>
       <array name="nic" optimistic-lock="true">
        <key on-delete="noaction" unique="true">
         <column name="dev_id" sql-type="int" not-null="false" length="0"/>
        </key>
        <list-index column="IndexColumnB"/>
        <one-to-many class="eb.nms.db.Nic"/>
       </array>
       <set name="disk" outer-join="false" inverse="true" lazy="true" optimistic-lock="true" cascade="save-update">
        <key on-delete="noaction" unique="true">
         <column name="dev_id" sql-type="int" not-null="false" length="0"/>
        </key>
        <one-to-many class="eb.nms.db.Disk"/>
       </set>
      </joined-subclass>


DISK的:
   <class name="Disk" table="disk" mutable="true" lazy="true" abstract="false">
      <id name="diskId">
         <column name="disk_id" sql-type="int" not-null="true"/>
         <generator class="native">  
         </generator>
      </id>
      <property name="diskName" insert="true" update="true" optimistic-lock="true">
         <column name="disk_name" sql-type="varchar(254)" length="254"/>
      </property>
      <many-to-one name="host" class="Host" outer-join="false" update="true" insert="true">
       <column name="dev_id" sql-type="int" not-null="false" length="0"/>
      </many-to-one>
   </class>


附文:
 

Hibernate Gossip: 雙向關聯(inverse 的意義)

多對一 一對多 中都是單向關聯,也就是其中一方關聯到另一方,而另一方不知道自己被關聯。

如果讓雙方都意識到另一方的存在,這就形成了雙向關聯,在多對一、一對多的例子可以改寫一下,重新設計User類別如下:

  • User.java

package onlyfun.caterpillar;

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

    public Integer getId() {
        return id;
    }

    public void setId(Integer 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
類別如下:

  • Room.java

package onlyfun.caterpillar;

import java.util.Set;

public class Room {
    private Integer id;
    private String address;
    private Set users;
 
    public Room() {}
 
    public Integer getId() {
        return id;
    }

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

    public Set getUsers() {
        return users;
    }

    public void setUsers(Set users) {
        this.users = users;
    }
 
    public void addUser(User user) {
        users.add(user);
    }
 
    public void removeUser(User user) {
        users.remove(user);
    }
}


如此,User實例可參考至Room實例而維持多對一關係,而Room實例記得User實例而維持一對多關係。

在映射文件方面,可以如下撰寫:

  • 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" type="java.lang.Integer">
            <generator class="native"/>
        </id>

        <property name="name" column="name" type="java.lang.String"/>
 
        <many-to-one name="room"
                     column="room_id"
                     class="onlyfun.caterpillar.Room"
                     cascade="save-update"
                     outer-join="true"/>
    </class>

</hibernate-mapping>

 

  • 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"
                  type="java.lang.String"/>
 
        <set name="users" table="user" cascade="save-update">
            <key column="room_id"/>
            <one-to-many class="onlyfun.caterpillar.User"/>
        </set>
    </class>

</hibernate-mapping>


映射文件雙方都設定了cascadesave-update,所以您可以用多對一的方式來維持關聯:

User user1 = new User();
user1.setName("bush");
       
User user2 = new User();
user2.setName("caterpillar");

Room room1 = new Room();
room1.setAddress("NTU-M8-419");

user1.setRoom(room1);
user2.setRoom(room1);
       
Session session = sessionFactory.openSession();
Transaction tx = session.beginTransaction();
       
session.save(user1);
session.save(user2);

tx.commit();
session.close();


或是反過來由一對多的方式來維持關聯:

User user1 = new User();
user1.setName("bush");
       
User user2 = new User();
user2.setName("caterpillar");

Room room1 = new Room();
room1.setUsers(new HashSet());
room1.setAddress("NTU-M8-419");
room1.addUser(user1);
room1.addUser(user2);
       
Session session = sessionFactory.openSession();
Transaction tx = session.beginTransaction();
       
session.save(room1);

tx.commit();
session.close();


這邊有個效率議題可以探討,上面的程式片段Hibernate將使用以下的SQL進行儲存:

Hibernate: insert into room (address) values (?)
Hibernate: insert into user (name, room_id) values (?, ?)
Hibernate: insert into user (name, room_id) values (?, ?)
Hibernate: update user set room_id=? where id=?
Hibernate: update user set room_id=? where id=?


上面的程式寫法表示關聯由Room單方面維持,而主控方也是RoomUser不知道Roomroom_id是多少,所以必須分別儲存Room User之後,再更新userroom_id

在一對多、多對一形成雙向關聯的情況下,可以將關聯維持的控制權交給多的一方,這樣會比較有效率,理由不難理解,就像是在公司中,老闆要記住多個員工的姓名快,還是每一個員工都記得老闆的姓名快。

所以在一對多、多對一形成雙向關聯的情況下,可以在「一」的一方設定控制權反轉,也就是當儲存「一」的一方時,將關聯維持的控制權交給「多」的一方,以上面的例子來說,可以設定Room.hbm.xml如下:

  • 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"
                  type="java.lang.String"/>
 
        <set name="users" table="user" cascade="save-update" inverse="true">
            <key column="room_id"/>
            <one-to-many class="onlyfun.caterpillar.User"/>
        </set>
    </class>

</hibernate-mapping>


由於關聯的控制權交給「多」的一方了,所以直接儲存「一」方前,「多」的一方必須意識到「一」的存在,所以程式片段必須改為如下:

User user1 = new User();
user1.setName("bush");
       
User user2 = new User();
user2.setName("caterpillar");

Room room1 = new Room();
room1.setUsers(new HashSet());
room1.setAddress("NTU-M8-419");
room1.addUser(user1);
room1.addUser(user2);

//
多方必須意識到單方的存在
user1.setRoom(room1);

user2.setRoom(room1);
       
Session session = sessionFactory.openSession();
Transaction tx = session.beginTransaction();
       
session.save(room1);

tx.commit();
session.close();


上面的程式片段Hibernate將使用以下的SQL

Hibernate: insert into room (address) values (?)
Hibernate: insert into user (name, room_id) values (?, ?)
Hibernate: insert into user (name, room_id) values (?, ?)


如果控制權交給另一方了,而另一方沒有意識到對方的存在的話會如何?試著將上面的程式片段中user1.setRoom(room1); user2.setRoom(room1);移去,執行之後,您會發現資料庫中room_id會出現null值,這種結果就好比在 多對一 中,您沒有分配給User一個Room,理所當然的,room_id會出現null

 




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posted on 2007-04-15 17:17 我爱佳娃 阅读(6083) 评论(0)  编辑  收藏 所属分类: Hibernate

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