使用Fluent NHibernate进行继承映射

Question

鉴于以下场景,我希望使用Fluent NHibernate将类型层次结构映射到数据库模式.

我正在使用NHibernate 2.0

---

键入层次结构

public abstract class Item
{
    public virtual int ItemId { get; set; }
    public virtual string ItemType { get; set; }
    public virtual string FieldA { get; set; }
}

public abstract class SubItem : Item
{
    public virtual string FieldB { get; set; } 
}

public class ConcreteItemX : SubItem
{
    public virtual string FieldC { get; set; } 
}

public class ConcreteItemY : Item
{
    public virtual string FieldD { get; set; }
}

见图

在Item和SubItem类是抽象的.

---

数据库架构

+----------+  +---------------+  +---------------+
| Item     |  | ConcreteItemX |  | ConcreteItemY |
+==========+  +===============+  +===============+
| ItemId   |  | ItemId        |  | ItemId        |
| ItemType |  | FieldC        |  | FieldD        |
| FieldA   |  +---------------+  +---------------+
| FieldB   |
+----------+

见图

该ItemType字段确定具体类型.

ConcreteItemX表中的每条记录在表中都有一条相应的记录Item; 同样适用于ConcreteItemY桌子.

FieldB如果项类型是,则始终为null ConcreteItemY.

---

映射(到目前为止)

public class ItemMap : ClassMap<Item>
{
    public ItemMap()
    {
        WithTable("Item");
        Id(x => x.ItemId, "ItemId");
        Map(x => x.FieldA, "FieldA");

        JoinedSubClass<ConcreteItemX>("ItemId", MapConcreteItemX);
        JoinedSubClass<ConcreteItemY>("ItemId", MapConcreteItemY);
    }

    private static void MapConcreteItemX(JoinedSubClassPart<ConcreteItemX> part)
    {
        part.WithTableName("ConcreteItemX");
        part.Map(x => x.FieldC, "FieldC");
    }

    private static void MapConcreteItemY(JoinedSubClassPart<ConcreteItemY> part)
    {
        part.WithTableName("ConcreteItemX");
        part.Map(x => x.FieldD, "FieldD");
    }
}

FieldB 未映射.

---

问题

如何使用Fluent NHibernate 映射类的FieldB属性SubItem？

有什么办法可以利用DiscriminateSubClassesOnColumn这个ItemType领域吗？

---

附录

我可以使用hbm.xml文件实现所需的结果:

<class name="Item" table="Item">

  <id name="ItemId" type="Int32" column="ItemId">
    <generator class="native"/>
  </id>

  <discriminator column="ItemType" type="string"/>

  <property name="FieldA" column="FieldA"/>

  <subclass name="ConcreteItemX" discriminator-value="ConcreteItemX">
    <!-- Note the FieldB mapping here -->
    <property name="FieldB" column="FieldB"/>
    <join table="ConcreteItemX">
      <key column="ItemId"/>
      <property name="FieldC" column="FieldC"/>
    </join>
  </subclass>

  <subclass name="ConcreteItemY" discriminator-value="ConcreteItemY">
    <join table="ConcreteItemY">
      <key column="ItemId"/>
      <property name="FieldD" column="FieldD"/>
    </join>
  </subclass>

</class>

如何使用Fluent NHibernate完成上述映射？

是否可以使用Fluent NHibernate将table-per-class-hierarchy与table-per-subclass混合使用？

Answer 1

我知道这已经很老了,但是现在设置流畅来生成你最初想要的精确映射非常简单.因为我在搜索答案时遇到了这个帖子,我以为我会发布它.

您只需为基类创建ClassMap,而无需引用您的子类:

public class ItemMap : ClassMap<Item>
{
    public ItemMap()
    {
        this.Table("Item");
        this.DiscriminateSubClassesOnColumn("ItemType");
        this.Id(x => x.ItemId, "ItemId");
        this.Map(x => x.FieldA, "FieldA");
    }
}

然后映射你的抽象子类,如下所示:

public class SubItemMap: SubclassMap<SubItemMap>
{
    public SubItemMap()
    {
        this.Map(x => x.FieldB);
    }
}

然后映射你的具体子类,如下所示:

public class ConcreteItemXMap : SubclassMap<ConcreteItemX>
{
    public ConcretItemXMap()
    {
        this.Join("ConcreteItemX", x =>
        {
            x.KeyColumn("ItemID");
            x.Map("FieldC")
        });
    }
}

希望这有助于其他人用流利的方式寻找这种类型的映射.

Answer 2

好吧，我不确定它是否完全正确，但它可能会起作用......如果有人可以更干净地做到这一点，我很乐意看到它（说真的，我会的；这是一个有趣的问题）。

使用您提供的确切类定义，以下是映射：

public class ItemMap : ClassMap<Item>
{
    public ItemMap()
    {
        Id(x => x.ItemId);
        Map(x => x.ItemType);
        Map(x => x.FieldA);

        AddPart(new ConcreteItemYMap());
    }
}

public class SubItemMap : ClassMap<SubItem>
{
    public SubItemMap()
    {
        WithTable("Item");

        // Get the base map and "inherit" the mapping parts
        ItemMap baseMap = new ItemMap();
        foreach (IMappingPart part in baseMap.Parts)
        {
            // Skip any sub class parts... yes this is ugly
            // Side note to anyone reading this that might know:
            // Can you use GetType().IsSubClassOf($GenericClass$)
            // without actually specifying the generic argument such
            // that it will return true for all subclasses, regardless
            // of the generic type?
            if (part.GetType().BaseType.Name == "JoinedSubClassPart`1")
                continue;
            AddPart(part);
        }
        Map(x => x.FieldB);
        AddPart(new ConcreteItemXMap());
    }
}

public class ConcreteItemXMap : JoinedSubClassPart<ConcreteItemX>
{
    public ConcreteItemXMap()
        : base("ItemId")
    {
        WithTableName("ConcreteItemX");
        Map(x => x.FieldC);
    }
}

public class ConcreteItemYMap : JoinedSubClassPart<ConcreteItemY>
{
    public ConcreteItemYMap()
        : base("ItemId")
    {
        WithTableName("ConcreteItemY");
        Map(x => x.FieldD);
    }
}

这些映射生成两个 hbm.xml 文件，如下所示（为了清楚起见，删除了一些无关的数据）：

  <class name="Item" table="`Item`">
    <id name="ItemId" column="ItemId" type="Int32">
      <generator class="identity" />
    </id>
    <property name="FieldA" type="String">
      <column name="FieldA" />
    </property>
    <property name="ItemType" type="String">
      <column name="ItemType" />
    </property>
    <joined-subclass name="ConcreteItemY" table="ConcreteItemY">
      <key column="ItemId" />
      <property name="FieldD">
        <column name="FieldD" />
      </property>
    </joined-subclass>
  </class>

  <class name="SubItem" table="Item">
    <id name="ItemId" column="ItemId" type="Int32">
      <generator class="identity" />
    </id>
    <property name="FieldB" type="String">
      <column name="FieldB" />
    </property>
    <property name="ItemType" type="String">
      <column name="ItemType" />
    </property>
    <property name="FieldA" type="String">
      <column name="FieldA" />
    </property>
    <joined-subclass name="ConcreteItemX" table="ConcreteItemX">
      <key column="ItemId" />
      <property name="FieldC">
        <column name="FieldC" />
      </property>
    </joined-subclass>
  </class>

它很丑陋，但看起来它可以生成一个可用的映射文件，而且它很流畅！：/您也许可以进一步调整这个想法，以获得您想要的结果。