我有以下对象模型:
public class Organizer
{
private int id;
public virtual int Id
{
get { return id; }
set { id = value; }
}
private string fullName = "";
public virtual string FullName
{
get { return fullName; }
set { fullName = value; }
}
private List<Email> emails = new List<Email>();
public virtual List<Email> Emails
{
get { return emails; }
set { emails = value; }
}
}
public enum EmailType
{
Primary, Secondary
}
public class Email
{
private int iD;
public virtual int ID
{
get { return iD; }
set { iD = value; }
}
private string emailAddress = "";
public virtual string EmailAddress
{
get { return emailAddress; }
set { emailAddress = value; }
}
private EmailType emailType = EmailType.Primary;
public virtual EmailType EmailType
{
get { return emailType; }
set { emailType = value; }
}
private List<Organizer> organizers;
public virtual List<Organizer> Organizers
{
get { return organizers; }
set { organizers = value; }
}
}
我有以下数据库架构:
CREATE TABLE [dbo].[EmailAddresses](
[ID] [int] IDENTITY(1,1) NOT NULL,
[EmailAddress] [nvarchar](550) NULL
)
CREATE TABLE [dbo].[Organizers](
[ID] [int] IDENTITY(1,1) NOT NULL,
[FullName] [nvarchar](550) NULL
)
CREATE TABLE [dbo].[Organizers_PrimaryKeys](
[ID] [int] IDENTITY(1,1) NOT NULL,
[PrimaryKeyID] [int] NULL,
[PrimaryKeyTypeID] [int] NULL,
[OrganizerID] [int] NULL
)
现在,我遇到了严重的问题,想弄清楚如何用nhibernate(不流利的nhibernate)来映射它。基本上,我希望它映射为当我执行“session.Save(anOrganizer)”时,数据有时会写入两个表,有时会写入三个表。例如,假设我们有OrganizerA,他有三封电子邮件:email_1,email_2和email_3。 email_1有EmailType.Primary,而另外两个有EmailType.Secondary。因此,当我们保存OrganizerA时,会发生以下情况:
那么我如何映射这种关系呢?到目前为止,我有以下内容:
<?xml version="1.0" encoding="utf-8" ?>
<hibernate-mapping xmlns="urn:nhibernate-mapping-2.2"
assembly="BLL"
namespace="BusinessLogic">
<class name="Organizer" table="Organizers">
<id name="Id">
<column name="ID"/>
<generator class="native" />
</id>
<property name="FullName">
<column name="FullName"/>
</property>
<bag name="Emails" table="Organizers_PrimaryKeys" inverse="false" cascade="all" lazy="true" >
<key column="Id"/>
<many-to-many class="Email" />
</bag>
</class>
</hibernate-mapping>
<?xml version="1.0" encoding="utf-8" ?>
<hibernate-mapping xmlns="urn:nhibernate-mapping-2.2"
assembly="BLL"
namespace="BusinessLogic">
<class name="Email" table="EmailAddresses">
<id name="ID">
<column name="ID"/>
<generator class="native" />
</id>
<property name="EmailAddress" unique="true" type="string" >
<column name="EmailAddress"/>
</property>
</class>
</hibernate-mapping>
我不知道在Email.hbm.xml中放什么来完成这种“有条件的”多对多关系。
答案 0 :(得分:2)
您需要将主要和次要电子邮件分隔为两个不同的对象,因为它们不是同一类型的实体。添加Organizer类与PrimaryEmail的一对一关系,并保留当前的多对多SecondaryEmail列表。
<?xml version="1.0" encoding="utf-8" ?>
<hibernate-mapping xmlns="urn:nhibernate-mapping-2.2"
assembly="BLL"
namespace="BusinessLogic">
<class name="Organizer" table="Organizers">
<id name="Id">
<column name="ID"/>
<generator class="native" />
</id>
<property name="FullName">
<column name="FullName"/>
</property>
<one-to-one name="PrimaryEmail" class="PrimaryEmail" />
<bag name="Emails" table="EmailAddresses" inverse="false" cascade="all" lazy="true" >
<key column="Id"/>
<many-to-many class="SecondaryEmail" />
</bag>
</class>
<class name="PrimaryEmail" table="Organizers_PrimaryKeys">
<id name="ID">
<column name="ID"/>
<generator class="native" />
</id>
<property name="EmailAddress" unique="true" type="string" >
<column name="EmailAddress"/>
</property>
</class>
<class name="SecondaryEmail" table="EmailAddresses">
<id name="ID">
<column name="ID"/>
<generator class="native" />
</id>
<property name="EmailAddress" unique="true" type="string" >
<column name="EmailAddress"/>
</property>
</class>
答案 1 :(得分:0)
email_1,email_2和email_3作为属性
附加到Organizers表