为什么我的多级继承映射不​​能像我期望的那样工作？

Question

我有以下实体映射：

// user.cfc
component persistent="true" table="user" discriminatorColumn="userTypeID" {

    property name="id" column="userID" fieldtype="id" generator="identity";
    property name="type" fieldtype="many-to-one" cfc="userType" fkcolumn="userTypeID";

}

// admin.cfc
component extends="user" persistent="true" table="admin" joincolumn="userID" discriminatorValue="3" {

    property name="id" column="adminID" fieldtype="id" generator="identity";

}

// employee.cfc
component extends="user" persistent="true" table="employee" joincolumn="userID" discriminatorValue="0" {

    property name="id" column="employeeID" fieldtype="id" generator="identity";

}

// manager.cfc
component extends="employee" persistent="true" table="manager" joincolumn="employeeID" discriminatorValue="1" {

    property name="id" column="managerID" fieldtype="id" generator="identity";

}

// intern.cfc
component extends="employee" persistent="true" table="intern" joincolumn="employeeID" discriminatorValue="2" {

    property name="id" column="internID" fieldtype="id" generator="identity";

}

Per Henry的建议，这里是生成的hbmxml文件：

<!-- user.hbmxml -->
<!DOCTYPE hibernate-mapping PUBLIC "-//Hibernate/Hibernate Mapping DTD 3.0//EN"
                                   "http://hibernate.sourceforge.net/hibernate-mapping-3.0.dtd">
<hibernate-mapping>
    <class entity-name="user" lazy="true"
        name="cfc:user" table="user">
        <id name="ID" type="int">
            <column name="userID"/>
            <generator class="identity"/>
        </id>
        <discriminator column="userTypeID"/>
        <many-to-one class="cfc:userType"
            column="userTypeID" insert="false" name="type" update="false"/>
    </class>
</hibernate-mapping>


<!-- admin.hbmxml -->
<!DOCTYPE hibernate-mapping PUBLIC "-//Hibernate/Hibernate Mapping DTD 3.0//EN"
                                   "http://hibernate.sourceforge.net/hibernate-mapping-3.0.dtd">
<hibernate-mapping>
    <subclass discriminator-value="3" entity-name="admin"
        extends="cfc:user" lazy="true" name="cfc:admin">
        <join table="admin">
            <key column="userID"/>
        </join>
    </subclass>
</hibernate-mapping>


<!-- employee.hbmxml -->
<!DOCTYPE hibernate-mapping PUBLIC "-//Hibernate/Hibernate Mapping DTD 3.0//EN"
                                   "http://hibernate.sourceforge.net/hibernate-mapping-3.0.dtd">
<hibernate-mapping>
    <subclass discriminator-value="0" entity-name="employee"
        extends="cfc:user" lazy="true" name="cfc:employee">
        <join table="employee">
            <key column="userID"/>
        </join>
    </subclass>
</hibernate-mapping>


<!-- manager.hbmxml -->
<!DOCTYPE hibernate-mapping PUBLIC "-//Hibernate/Hibernate Mapping DTD 3.0//EN"
                                   "http://hibernate.sourceforge.net/hibernate-mapping-3.0.dtd">
<hibernate-mapping>
    <subclass discriminator-value="1" entity-name="manager"
        extends="cfc:employee" lazy="true" name="cfc:manager">
        <join table="manager">
            <key column="employeeID"/>
        </join>
    </subclass>
</hibernate-mapping>


<!-- intern.hbmxml -->
<!DOCTYPE hibernate-mapping PUBLIC "-//Hibernate/Hibernate Mapping DTD 3.0//EN"
                                   "http://hibernate.sourceforge.net/hibernate-mapping-3.0.dtd">
<hibernate-mapping>
    <subclass discriminator-value="2" entity-name="intern"
        extends="cfc:employee" lazy="true" name="cfc:intern">
        <join table="intern">
            <key column="employeeID"/>
        </join>
    </subclass>
</hibernate-mapping>

如果映射不清楚，则关系如下：

user
 |- admin
 |- employee
  |- manager
  |- intern

意图是type实体上的user属性由manager和intern实体的鉴别器值填充。 employee的构造函数中有代码可以防止它直接实例化，因此user将始终具有type。

从DB读取一些已存在的数据时，整个映射工作正常。但是，当我尝试插入新记录时，我遇到了问题。

假设所涉及的表已经填充了一些记录：

user
|---------------------|
| userID | userTypeID |
|---------------------|
| 1      | 1          |
| 2      | 2          |
| 3      | 2          |
| 4      | 3          |
|---------------------|

admin
|------------------|
| adminID | userID |
|------------------|
| 1       | 4      |
|------------------|

employee
|---------------------|
| employeeID | userID |
|---------------------|
| 1          | 1      |
| 2          | 2      |
| 3          | 3      |
|---------------------|

manager
|------------------------|
| managerID | employeeID |
|------------------------|
| 1         | 1          |
|------------------------|

intern
|-----------------------|
| internID | employeeID |
|-----------------------|
| 1        | 2          |
| 2        | 3          |
|-----------------------|

如果我要创建一个新的intern实体并保留它，我希望插入三条记录：

INSERT user ( userID, userTypeID ) VALUES ( 5, 2 )
INSERT employee ( employeeID, userID ) VALUES ( 4, 5 )
INSERT intern ( internID, employeeID ) VALUES ( 3, 4 )

但是，实际执行的SQL如下：

INSERT user ( userID, userTypeID ) VALUES ( 5, 2 )
INSERT employee ( employeeID, userID ) VALUES ( 4, 5 )
INSERT intern ( internID, employeeID ) VALUES ( 3, 5 ) -- using the new userID instead of the new employeeID

最后，实际问题：

为什么在插入intern时，是使用userID代替employeeID？好像Hibernate忽略joincolumn上的intern属性，只使用joincolumn中的employee。

Answer 1

Hibernate将使用User表中的discriminatorColumn构造一个“实习生”对象，因此它使用userID而不是employeeID。