爲什麼我的多級繼承映射不​​能像我期望的那樣工作？

Question

我有以下實體映射：

// 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"; 

} 

每亨利的建議，這裏是產生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> 

如果不是從映射清晰，關係如下：

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

意圖是對user實體上的type屬性進行填充manager和intern實體的鑑別器值。 employee的構造函數中存在代碼，使其不能直接實例化，所以user將始終有一個type。

從數據庫讀取一些已經存在的數據時，整個映射工作正常。但是，當我試圖插入新記錄時遇到問題。

假設已經涉及到的表有一些記錄填充：

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   | 
|-----------------------| 

如果我要創建一個新的intern實體和堅持它，我希望要插入三個記錄：

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

但是，實際執行的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 

最後，實際問題：

爲什麼在插入到intern時，是使用userID而不是employeeID？就好像Hibernate忽略intern上的joincolumn屬性並僅使用employee中的joincolumn。

Answer 1

Hibernate將使用User表中的discriminatorColumn構造一個「intern」對象，以便它使用userID而不是employeeID。