java当不是对象本身的属性时，如何在manytone关系上指定外键？

Question

我正在尝试将MarketMenuBranch和Market之间的多对一关系映射到一起。我的课看起来像：

public class Market implements Serializable {
    private int id;
    private String name;
    private List<MarketMenuBranch> marketMenuBranches;
    // accessors / mutators etc...

public class MarketMenuBranch implements Serializable {
    private MarketMenuBranchId id;
    private String name;
    // accessors / mutators etc...

public class MarketMenuBranchId implements Serializable {
    private int marketId;
    private int sequence;
   // accessors / mutators etc...

但我不知道我能为房产名称写些什么（我在下面有？？？）。我真的很想把id市场化，但这似乎是错误的

<class name="MarketMenuBranch" table="MARKET_MENU_BRANCH">
    <composite-id name="id" class="MarketMenuBranchId">
        <key-property name="marketId"/>
        <key-property name="sequence"/>
    </composite-id>
    <property name="name"/>
    <many-to-one name="????????"/>
</class>

我该怎么做

编辑：以下是市场地图：

<class name="Market" table="MARKET">
    <id name="id"/>
    <property name="name"/>
    <list name="marketMenuBranches" inverse="true" cascade="all-delete-orphan">
        <key column="marketId"/>
        <list-index column="sequence"/>
        <one-to-many class="MarketMenuBranch"/>
    </list>
</class>

这是我的桌子：

mysql> describe MARKET;
+-------+--------------+------+-----+---------+-------+
| Field | Type         | Null | Key | Default | Extra |
+-------+--------------+------+-----+---------+-------+
| id    | int(11)      | NO   | PRI | NULL    |       | 
| name  | varchar(100) | YES  |     | NULL    |       | 
+-------+--------------+------+-----+---------+-------+

mysql> describe MARKET_MENU_BRANCH;
+----------+--------------+------+-----+---------+-------+
| Field    | Type         | Null | Key | Default | Extra |
+----------+--------------+------+-----+---------+-------+
| marketId | int(11)      | NO   | PRI | 0       |       | 
| sequence | int(11)      | NO   | PRI | 0       |       | 
| name     | varchar(100) | YES  |     | NULL    |       | 
+----------+--------------+------+-----+---------+-------+

Answer 1

我认为你不应该在MarketMenuBranch中给你的数据成员命名为“id”，因为这有点让人困惑。我推荐marketMenuBranch。是这样吗

但是，离开上面的内容，你只是引用了你所指的类成员的名称。所以：

<many-to-one name="id" column="column_name" class="MarketMenuBranchId"/>

您只需指向所创建的容器对象，并将其用作类成员。MarketMenuBranchId也应该被映射，Hibernate将知道如何把id从那里取出来，这就是为什么你不需要把id.marketId放进去

顺便说一句：我不知道整个结构是否正确（这取决于你），但这应该能回答你的具体问题。对我来说，你给XXXId打电话的方式似乎有点误导，但你比我更清楚你的需求

编辑：我还想指出，如果你对数据库有任何控制权，我建议不要使用复合ID。这会让未来的事情变得有点麻烦。系统生成的id更好地遵循规范化规则

Answer 2

也许您应该将MarketMenuBranch映射为复合元素列表：

<class name="Market" table="MARKET">
    ...
    <list cascade="all-delete-orphan" name="marketMenuBranches" table="MARKET_MENU_BRANCH">
      <key column="MARKET_ID"/>
      <list-index column="SEQUENCE"/>
      <composite-element class="MarketMenuBranch">
        <property column="NAME" name="name"/>
      </composite-element>
    </list>

Answer 3

出现问题是因为有多个属性共享同一列。你的MarketMenuBranch应该是这样的（注意insert=“false”，update=“false”）

<class name="MarketMenuBranch" table="MARKET_MENU_BRANCH">
    <composite-id name="id" class="MarketMenuBranchId">
        <! Ops... A property whose mapped column is called marketId >
        <key-property name="marketId"/>
        <key-property name="sequence"/>
    </composite-id>
    <property name="name"/>
    <! Ops... Other property whose mapped column is called marketId >
    <many-to-one name="market" column="marketId" insert="false" update="false"/>
</class>

Hibernate会问你：我应该保存/更新哪个属性

因此，您应该定义insert=“false”，update=“false”

你是新来冬眠的吗

Keep this in mind: Hibernate does not support automatic generation of composite primary key

因此，在保存前，您应该提供复合主键