如何用共享列将组合id和多个id映射为一个(多列)

Question

在大型遗留数据库中，复合id和多列多对一关系之间的共享列存在问题。我将问题简化为映射，下面是问题的再现：

<?xml version="1.0" encoding="utf-8" ?>
<hibernate-mapping xmlns="urn:nhibernate-mapping-2.2">
    <class name="Alfa">    
    <composite-id name="AlfaId" class="AlfaCompositeId">
      <key-many-to-one name="Bravo" class="Bravo" column="BRAVO_ID" />
      <key-property name="CharlieId1" column="CHARLIE_ID1" />
    </composite-id>
    <many-to-one name="C" class="Charlie">
      <column name="CHARLIE_ID1" />
      <column name="BRAVO_ID" />
    </many-to-one>
    <property name="CookieA" column="COOKIE_A" />
    </class>
  <class name="Bravo">
    <id name="BravoId" column="BRAVO_ID" />
    <property name="CookieB" column="COOKIE_B" />
  </class>
  <class name="Charlie">
    <composite-id>
      <key-property name="CharlieId1" column="CHARLIE_ID1" />
      <key-many-to-one name="Bravo" class="Bravo" column="BRAVO_ID" /> <!-- Reused column !!! -->
    </composite-id>
    <property name="CookieC" />
  </class>
</hibernate-mapping>

试图保存Alfa实体将引发：

IndexOutOfRangeException (Message=Invalid index 3 for this SqlParameterCollection with Count=3).

NHibernate试图两次添加BRAVO_ID列值。一次用于复合id，一次用于多到一个FK。

如何更改为映射XML以告诉NHibernate忽略'BRAVO_ID‘列的第二个映射？

任何小费都会得到赏识。

Answer 1

您将相同的列映射为CompositeId和ManyToOne，只需去掉compositeId

<?xml version="1.0" encoding="utf-8" ?>
<hibernate-mapping xmlns="urn:nhibernate-mapping-2.2">
    <class name="Alfa">    
    <composite-id>
      <key-many-to-one name="Charly" class="Charlie">
        <column name="CHARLIE_ID1" />
        <column name="BRAVO_ID" />
      </key-many-to-one>
    </composite-id>
    <property name="CookieA" column="COOKIE_A" />
  </class>

  <class name="Bravo">
    <id name="Id" column="BRAVO_ID" />
    <property name="CookieB" column="COOKIE_B" />
  </class>

  <class name="Charlie">
    <composite-id>
      <key-property name="IdPart1" column="CHARLIE_ID1" />
      <key-many-to-one name="Bravo" class="Bravo" column="BRAVO_ID" />
    </composite-id>
    <property name="CookieC" />
  </class>
</hibernate-mapping>

并通过该属性访问它

Bavo = alfa.Charly.Bravo；