映射同一个类关系

一尘不染

映射同一个类关系

hibernate

嗨，我正在尝试在hibernate状态下映射一些类，但在如何完成此类映射方面存在一般性问题。有User类和Facebook用户类，它们具有以下结构User Class：

public class User{
 public User(){}
 Long Id;
 String FirstName;
 String LastName;
 ....
 FbUser fbuser;
 //// all requred 
 getters and setters...
}

Facebook类FbUser可以具有属于同一类FbUser的对象的Friends列表。

public class FbUser{
 public FbUser(){}
 Long fbId;
 String FirstName;
 String LastName;
 List<FbUser> friends;
 //// all requred 
 getters and setters...
}

到现在为止，我在User和FbUser之间建立了许多一对一的关系。

<hibernate-mapping>
    <class
        name="User"
        table="User"
    >

        <id
            name="Id"
            column="ID"
            type="java.lang.Long"
            unsaved-value="null"
        >
         <generator class="increment"/>
        </id>

        <property
            name="FirstName"
            update="true"
            insert="true"
            not-null="false"
            unique="false"
            type="java.lang.String"
        >
            <column name="FirstName" />
        </property>
 <property
            name="LastName"
            update="true"
            insert="true"
            not-null="false"
            unique="false"
            type="java.lang.String"
        >
            <column name="LastName" />
        </property>
        <many-to-one
            name="fbUser"
            class="FbUser"
            cascade="all"
            column="fbId"
            unique="true" 
        />

    </class>
</hibernate-mapping>

现在，FbUser映射：

<hibernate-mapping>
    <class
        name="FbUser"
        table="FbUser"
    >

        <id
            name="fbId"
            column="fbId"
            type="java.lang.Long"
            unsaved-value="null"
        >
          <generator class="increment"/>
        </id>

        <property
            name="FirstName"
            update="true"
            insert="true"
            not-null="false"
            unique="false"
            type="java.lang.String"
        >
            <column name="FirstName" />
        </property>

        <property
            name="LastName"
            type="java.lang.String"
            update="true"
            insert="true"
            column="LastName"
            not-null="true"
            unique="false"
        />
    </class>
</hibernate-mapping>

如何在FbUser Map文件中映射FbUser List？我迷路了：（

阅读 241

2020-06-20

共1个答案

一尘不染

您可以创建一个名为MyFriends的附加类。

public class FbUser {

    List<MyFriends> friends = new ArrayList<MyFriends>();

}

只是相关的部分

如果您有索引栏

<hibernate-mapping>
    <class name="FbUser">
        <list name="myFriends">
            <key column="ME_ID" insert="false" update="false"/>
            <list-index column="WHICH COLUMN SHOULD BE USED AS INDEX"/>
            <one-to-many class="MyFriends"/>
        </list>
    </class>
</hibernate-mapping>

如果您没有索引列

将您的清单改写为

public class FbUser {

    Collection<MyFriends> friends = new ArrayList<MyFriends>();

}

和

<hibernate-mapping>
    <class name="FbUser">
        <bag name="columns">
            <key column="ME_ID" insert="false" update="false"/>
            <one-to-many class="MyFriends"/>
        </bag>
    </class>
</hibernate-mapping>

以及您的MyFriends映射。 注意，您需要一个复合主键 （实现为静态内部类）

<class name="MyFriends">
    <composite-id name="myFriendsId" class="MyFriends$MyFriendsId">
        <key-property name="meId"/>
        <key-property name="myFriendId"/>
    </composite-id>
    <many-to-one name="me" class="FbUser" insert="false" update="false"/>
    <many-to-one name="myFriend" class="FbUser" insert="false" update="false"/>
</class>

您的MyFriends如下所示

public class MyFriends {

    private MyFriendsId myFrinedId;

    private FbUser me;
    private FbUser myFriend;

    public static class MyFriendsId implements Serializable {

        private Integer meId;
        private Integer myFriendId;

        // getter's and setter's

        public MyFriendsId() {}
        public MyFriendsId(Integer meId, Integer myFriendId) {
            this.meId = meId;
            this.myFriendId = myFriendId;
        }

        // getter's and setter's

        public boolean equals(Object o) {
            if(!(o instanceof MyFriendsId))
                return false;

            MyFriendsId other = (MyFriendsId) o;
            return new EqualsBuilder()
                       .append(getMeId(), other.getMeId())
                       .append(getMyFriendId(), other.getMyFriendId())
                       .isEquals();
        }

        public int hashcode() {
            return new HashCodeBuilder()
                       .append(getMeId())
                       .append(getMyFriendId())
                       .hashCode();
        }
    }
}

2020-06-20