Question

我有一个查询，当我在sqlDeveloper中运行时，我得到了结果。这是查询

select * from Losa_Cust_Reject_App rejectApp 
inner join 
    Losa_Cust losaCust 
on 
    rejectApp.app_Ref_No = losaCust.app_Ref_No 
where 
    rejectApp.app_Ref_No != '0000001912' and  rejectApp.app_Dt >= '23-SEP-2012' 
and
    rejectApp.cust_Id1 = 'A111111111' and rejectApp.cust_Id1_Type = '01';

我将此查询翻译为hibernate，如下所示

public List findAppByDate(String appRefNo, Date previousAppDate, String currentId1,
    String currentIdType1) {

    StringBuffer query = new StringBuffer("from ");
    query.append(LosaCustRejectApp.class.getName());
    query.append(" rejectApp inner join ");
    query.append(LosaCust.class.getName());
    query.append(" losaCust with rejectApp.appRefNo = losaCust.comp_id.appRefNo");
    query.append(" where rejectApp.appRefNo != ? and  rejectApp.appDt >= ?");
    query.append(" and rejectApp.custId1 = ? and rejectApp.custId1Type = ? ");

    List result = null;

    try {
        result = getHibernateTemplate().find(query.toString(),
            new Object[] { appRefNo, previousAppDate, currentId1, currentIdType1 });

        if (CollectionUtils.isNotEmpty(result)) {
            return result;
        }
    } catch (Exception e) {
        String message = e.getMessage();
        System.out.println();
    }
    return result;
}

但是我想在hibernate中定义映射，所以hibernate会自动进行连接。如何在hibernate文件中为上述查询定义关联，以便hibernate自动进行连接。

这是我的LosaCustReject.hbm.xml

<hibernate-mapping default-lazy="false">
<class name="com.thetasp.losa.data.LosaCustRejectApp" table="LOSA_CUST_REJECT_APP" optimistic-lock="version">

    <id name="rejectAppId" type="java.lang.Long" column="REJECT_APP_ID">
        <generator class="com.thetasp.code.runningno.HibernateId">
            <param name="fieldCode">REJECT_APP_ID</param>
            <param name="returnType">java.lang.Long</param>
        </generator>
    </id>
    <version name="rowVersion" type="int" column="ROW_VERSION" access="property" unsaved-value="undefined" />
    <property name="appRefNo" type="java.lang.String" column="APP_REF_NO" not-null="true" length="20" />
    <property name="appDt" type="java.sql.Timestamp" column="APP_DT" not-null="true" length="23" />
    <property name="custId1Type" type="java.lang.String" column="CUST_ID1_TYPE" not-null="true" length="1" />
    <property name="custId1" type="java.lang.String" column="CUST_ID1" not-null="true" length="20" />
    .......
    <property name="updateDt" type="java.sql.Timestamp" column="UPDATE_DT" length="23" />

    <!-- associations -->
</class>
</hibernate-mapping>

LosaCust.hbm.xml

<hibernate-mapping default-lazy="false">
<class name="com.thetasp.losa.data.LosaCust" table="LOSA_CUST" discriminator-value="I" optimistic-lock="version">

    <composite-id name="comp_id" class="com.thetasp.losa.data.LosaCustPK">
        <key-property name="custId" column="CUST_ID" type="java.lang.String" length="20" />
        <key-property name="appRefNo" column="APP_REF_NO" type="java.lang.String" length="20" />
    </composite-id>

    <discriminator column="CUST_T" type="string" length="1" >
    </discriminator>

    <version name="rowVersion" type="int" column="ROW_VERSION" access="property" unsaved-value="undefined" />
    <property name="custLevel" type="java.lang.String" column="CUST_LEVEL" not-null="false" length="10" />
              .....
    <property name="staySinceMth" type="java.lang.String" column="STAYED_SINCE_MTH" length="10" />
            ....
    <subclass name="com.thetasp.losa.data.LosaIndvCust" discriminator-value="I" lazy="false" >
        <property name="id1Type" type="java.lang.String" column="ID1_TYPE" length="1" />
        <property name="id1" type="java.lang.String" column="ID1" length="20" />
         ....

 </subclass>

</class>

谢谢

Answer 1

我认为你可以做这样的事情

在LosaCust.java中定义一个字段，表示与LosaCustRejectApp实体的一对多关联：

List<LosaCustRejectApp> associatedLosaCustRejectApps;

在LosaCust.hbm.xml中，您可以使用以下内容映射此新字段：

<list name="associatedLosaCustRejectApps" table="LOSA_CUST_REJECT_APP" >
    <key>
        <column name="app_Ref_No" not-null="true" />
    </key>
    <one-to-many class="com.thetasp.losa.data.LosaCustRejectApp" />
</set>

有关更多详细信息，甚至是关于如何进行此类映射的完整教程，您可以访问：

Hibernate – One-to-Many example (XML Mapping)

祝你好运

如何在hibernate文件中定义关联

1 个答案: