Question

我对NHibernate有些新手，我遇到了一个试图加入两个表的问题。我有一个名称表和地址表。无论是否为地址记录返回任何结果，我都要提取名称记录。如果我在下面的代码中有一个地址记录工作，但一旦地址记录被删除我不再收到名称记录。我正在尝试这个（NHibernate Left Outer Join），但它对我不起作用。有什么想法吗？

映射：

<?xml version="1.0" encoding="utf-8" ?>
<hibernate-mapping 
    xmlns="urn:nhibernate-mapping-2.2" 
    namespace="Portlet.IncomingStudentInfo.Data.BusinessObjects" 
    assembly="Portlet.IncomingStudentInfo">
    <class name="ISINameMasterRecord" table="NAME_MASTER">
         <id name="NM_ID_NUM" column="ID_NUM" type="Int32">
            <generator class="native" />
        </id>
        <property name="NM_ID_NUM" column="ID_NUM" />
        <property name="NM_EMAIL_ADDRESS" column="EMAIL_ADDRESS" />
        <property name="NM_MOBILE_PHONE" column="MOBILE_PHONE" />
        <many-to-one name="LHP" class="ISILHPAddress" 
            column="ID_NUM" fetch="join" 
            foreign-key="ID_NUM" 
            outer-join="true" not-found="ignore" />
    </class>

    <class name="ISILHPAddress" table="ADDRESS_MASTER">
        <composite-id>
            <key-property name="AD_ID_NUM" column="ID_NUM" type="Int32" />
        </composite-id>
        <property name="AD_ID_NUM" column="ID_NUM" />
        <property name="AD_ADDR_CDE" column="ADDR_CDE" />
        <property name="AD_ADDRESS" column="ADDR_LINE_1" />
        <property name="AD_CITY" column="CITY" />
        <property name="AD_STATE" column="STATE" />
        <property name="AD_ZIP" column="ZIP" />
        <property name="AD_PHONE" column="PHONE" />
    </class>   
</hibernate-mapping>

门面：

public class ISINameMasterRecordFacade : JICSBaseFacade<ISINameMasterRecord>
{
    public ISINameMasterRecord FindIDCriteria(int id)
    {
        ICriteria criteria = this.CreateCriteria();
        criteria.Add(Expression.Eq("NM_ID_NUM", id));

        criteria.CreateAlias(
            "LHP", 
            "lhp", 
            NHibernate.SqlCommand.JoinType.LeftOuterJoin);
        criteria.Add(
            Expression.Or(
                Expression.IsNull("lhp.AD_ADDR_CDE"), 
                Expression.Eq("lhp.AD_ADDR_CDE", "*LHP")));


        return criteria.UniqueResult<ISINameMasterRecord>();
    }
}

Answer 1

感谢大家的帮助。我正在与另一位开发人员交谈，他提供了以下答案，这确实有效。以下是有关该问题的一些其他信息。 NAME_MASTER表保存记录的名称信息，主键为ID_NUM。 ADDRESS_MASTER表包含记录的所有地址，主键为ID_NUM和ADDR_CDE（它们的地址类型为：合法归属永久物，电子邮件，夏季地址等）记录可能没有* LHP（合法归属永久物）地址），但可能有其他地址记录。无论是否存在* LHP地址记录，我们都想拉出NAME_MASTER记录，因此这实际上是连接的另一个条件。以下工作正常，无论ADDRESS_MASTER中是否存在* LHP记录，都会提取NAME_MASTER记录。

<强> mappings.hbm.xml

<?xml version="1.0" encoding="utf-8" ?>
<hibernate-mapping xmlns="urn:nhibernate-mapping-2.2" namespace="Portlet.IncomingStudentInfo.Data.BusinessObjects" assembly="Portlet.IncomingStudentInfo">
  <class name="ISINameMasterRecord" table="NAME_MASTER">
    <id name="NM_ID_NUM" column="ID_NUM" type="Int32">
      <generator class="native" />
    </id>
    <property name="NM_ID_NUM" column="ID_NUM" />
    <property name="NM_EMAIL_ADDRESS" column="EMAIL_ADDRESS" />
    <property name="NM_MOBILE_PHONE" column="MOBILE_PHONE" />

    <bag name="Addresses" cascade="all" where="ADDR_CDE='*LHP'" lazy="false" fetch="join">
      <key column="ID_NUM"/>
      <one-to-many class="ISILHPAddress"/>
    </bag>
  </class>


  <class name="ISILHPAddress" table="ADDRESS_MASTER" lazy="false">
    <id name="AD_ID_NUM" column="ID_NUM" type="Int32">
      <generator class="native" />
    </id>
    <property name="AD_ID_NUM" column="ID_NUM" />
    <property name="AD_ADDR_CDE" column="ADDR_CDE" />

    <property name="AD_ADDRESS" column="ADDR_LINE_1" />
    <property name="AD_CITY" column="CITY" />
    <property name="AD_STATE" column="STATE" />
    <property name="AD_ZIP" column="ZIP" />
    <property name="AD_PHONE" column="PHONE" />
  </class>

</hibernate-mapping>

<强> ISINameMasterRecordFacade.cs

public ISINameMasterRecord FindIDCriteria(int id)
        {
            ICriteria criteria = this.CreateCriteria();
            criteria.Add(Expression.Eq("NM_ID_NUM", id));

            return criteria.UniqueResult<ISINameMasterRecord>();
        }

<强> ISINameMasterRecord.cs

public class ISINameMasterRecord : EXBase
    {
        public virtual int NM_ID_NUM { get; set; }
        public virtual String NM_EMAIL_ADDRESS { get; set; }
        public virtual String NM_MOBILE_PHONE { get; set; }

        public virtual ISILHPAddress LHP
        {
            get { return Addresses != null && Addresses.Any() ? Addresses[0] : null; }
            set
            {
                if (Addresses == null)
                    Addresses = new List<ISILHPAddress>();

                if (Addresses.Any())
                    Addresses[0] = value;
                else
                    Addresses.Add(value);
            }
        }

        //for mapping purposes
        protected virtual IList<ISILHPAddress> Addresses { get; set; }


        public ISINameMasterRecord()
        {

        }
    }

<强> ISILHPAddress.cs

public class ISILHPAddress : EXBase
    {
        public virtual int AD_ID_NUM { get; set; }
        public virtual String AD_ADDR_CDE { get; set; }
        public virtual String AD_ADDRESS { get; set; }
        public virtual String AD_CITY { get; set; }
        public virtual String AD_STATE { get; set; }
        public virtual String AD_ZIP { get; set; }
        public virtual String AD_PHONE { get; set; }

        public ISILHPAddress() { }
    }

Answer 2

在映射中使用fetch="join"将导致使用左外连接急切加载地址。我认为outer-join="true"不是有效属性。如果外键无效（即，如果ID_NUM存在但地址表中不存在记录），not-found="ignore"会阻止NHibernate抛出异常。

因为您已将fetch="join"设置为急切加载地址，

ISession.Get<ISINameMasterRecord>(id);

将使用左外连接返回对象和地址。如果地址不存在，地址将为null。

可能导致问题的一件事是您已将AD_ID_NUM映射为复合键。这似乎没有必要。

NHibernate左外连接名称到地址

2 个答案: