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"; 

} 

每亨利的建议，这里是产生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> 

如果不是从映射清晰，关系如下：

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

意图是对user实体上的type属性进行填充manager和intern实体的鉴别器值。 employee的构造函数中存在代码，使其不能直接实例化，所以user将始终有一个type。

从数据库读取一些已经存在的数据时，整个映射工作正常。但是，当我试图插入新记录时遇到问题。

假设已经涉及到的表有一些记录填充：

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实体和坚持它，我希望要插入三个记录：

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如下所示：

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：

为什么在插入到intern时，是使用userID而不是employeeID？就好像Hibernate忽略intern上的joincolumn属性并仅使用employee中的joincolumn。

Answer 1

Hibernate将使用User表中的discriminatorColumn构造一个“intern”对象，以便它使用userID而不是employeeID。