2

我有一个在概念上类似于此的类层次结构：

在此处输入图像描述

也就是说，有一个抽象基类 ( Relation) 和几个派生类。在实践中，Customer共享Supplier了很多代码，所以我将共同点重构为一个抽象类BusinessContact。现在实际的类层次结构如下所示：

在此处输入图像描述

或在代码中：

public abstract class Relation
{
    public virtual int Id { get; set; }
}

public class ContactPerson : Relation
{
    public virtual string PhoneNumber { get; set; }
}

public abstract class BusinessContact : Relation
{
    public virtual string Name { get; set; }
}

public class Customer : BusinessContact
{
    public virtual string CustomerNumber { get; set; } 
}

public class Supplier : BusinessContact
{
    public virtual string SupplierNumber { get; set; }
}

我想使用 NHibernate 中的连接子类，使用按代码映射 ( )将此层次结构映射到四个表 ( Relation、ContactPerson和)。我的映射如下所示：CustomerSupplierModelMapper

var mapper = new ModelMapper();

mapper.Class<Relation>(map =>
{
    map.Id(x => x.Id, id => id.Generator(Generators.Native));
});

mapper.JoinedSubclass<ContactPerson>(map =>
{
    map.Key(key => key.Column("Id"));
    map.Property(x => x.PhoneNumber);
});

mapper.JoinedSubclass<Customer>(map =>
{
    map.Key(key => key.Column("Id"));
    map.Property(x => x.Name);
    map.Property(x => x.CustomerNumber);
});

mapper.JoinedSubclass<Supplier>(map =>
{
    map.Key(key => key.Column("Id"));
    map.Property(x => x.Name);
    map.Property(x => x.SupplierNumber);
});

但是，一旦我尝试将映射添加到配置中，就会出现异常：

NHibernate.MappingException: Cannot extend unmapped class: BusinessContact

我基本上明白为什么会发生这种情况。生成的映射如下所示：

<?xml version="1.0" encoding="utf-8"?>
<hibernate-mapping xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema" namespace="JoinedSubClassMapping" assembly="JoinedSubClassMapping" xmlns="urn:nhib    ernate-mapping-2.2">
  <class name="Relation" abstract="true">
    <id name="Id" type="Int32">
      <generator class="native" />
    </id>
  </class>
  <joined-subclass name="ContactPerson" extends="Relation">
    <key column="Id" />
    <property name="PhoneNumber" />
  </joined-subclass>
  <joined-subclass name="Customer" extends="BusinessContact">
    <key column="Id" />
    <property name="CustomerNumber" />
  </joined-subclass>
  <joined-subclass name="Supplier" extends="BusinessContact">
    <key column="Id" />
    <property name="SupplierNumber" />
  </joined-subclass>
</hibernate-mapping>

Customer并在他们的属性中Supplier定义，就好像模型中的“正常”实体一样。由于没有 的映射，因此失败，或者当然。请注意，“名称”属性（定义在其中也不会出现在映射中。BusinessContactextendsBusinessContactBusinessContactBusinessContact

我希望映射看起来像这样：

<?xml version="1.0" encoding="utf-8" ?>
<hibernate-mapping xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema" namespace="JoinedSubClassMapping" assembly="JoinedSubClassMapping" xmlns="urn:nhibernate-mapping-2.2">
  <class name="Relation" abstract="true">
    <id name="Id" type="Int32">
      <generator class="native" />
    </id>
  </class>
  <joined-subclass name="ContactPerson" extends="Relation">
    <key column="Id" />
    <property name="PhoneNumber" />
  </joined-subclass>
  <joined-subclass name="Customer" extends="Relation">
    <key column="Id" />
    <property name="CustomerNumber" />
    <property name="Name" />
  </joined-subclass>
  <joined-subclass name="Supplier" extends="Relation">
    <key column="Id" />
    <property name="SupplierNumber" />
    <property name="Name" />
  </joined-subclass>
</hibernate-mapping>

也就是说，创建Supplier和Customer扩展 Relation 并包含（否则未映射的）BusinessContact类的所有映射属性。

我怎样才能做到这一点？

4

1 回答 1

2

这是一个解决方案：

class MyInspector : ExplicitlyDeclaredModel {
    public override bool IsEntity(Type type) {
        if (type == typeof (BusinessContact))
            return false;
        return base.IsEntity(type);
    }
}

var mapper = new ModelMapper(new MyInspector());
于 2013-03-20T19:14:30.340 回答