在hibernate中从多个表中检索数据
在hibernate中从多个表中检索数据
提问于 2012-04-14 03:41:48
我有两个表，即订阅者和联系人。表看起来像这样：
subscriber -> id, contact_id //contact_id is a foreign key
contact -> id, firstName, lastName, email, contactType我的Contact.hbm.xml文件如下所示：
<hibernate-mapping>
<class name="com.DBNAME.model.Contact" table="contact" >
<id name="id" type="int">
<column name="id" />
<generator class="identity" />
</id>
<property name="contactType" type="int">
<column name="contactType" sql-type="TINYINT"></column>
</property>
<property name="firstName" type="string">
<column name="firstName"></column>
</property>
<property name="lastName" type="string">
<column name="lastName"></column>
</property>
</class>
</hibernate-mapping>我的Subscriber.hbm.xml文件看起来像这样：
<hibernate-mapping>
<class name="com.DBNAME.model.Subscriber" table="subscriber" >
<id name="id" type="int">
<column name="id" />
<generator class="identity" />
</id>
<many-to-one name="contact" class="com.DBNAME.model.Contact" column="contact_id" unique="true" fetch="join"/>
</class>
</hibernate-mapping>现在，我想检索一个简单的Subscriber对象，在该对象中自动映射联系人。因此，我在Java代码中所做的是：
/**
* get Subscribers
*/
@SuppressWarnings("unchecked")
private void getSubscribersWithContactDetails() {
Session session = HibernateUtils.getSessionFactory().getCurrentSession();
session.beginTransaction();
try {
setSubscribers((List<Subscriber>)session.createQuery("from Subscriber").list());
} catch (HibernateException e) {
session.getTransaction().rollback();
} finally {
session.getTransaction().commit();
}
}
/**
* @param subscribers the subscribers to set
*/
public void setSubscribers(List<Subscriber> subscribers) {
this.subscribers = subscribers;
}我的数据类如下所示：
public class Contact implements Serializable {
private static final long serialVersionUID = 1L;
private int id;
private int contactType;
private String firstName;
private String lastName;
// Getters Setters and constructors
}
public class Subscriber implements Serializable {
private static final long serialVersionUID = 1L;
private int id;
private Contact contact; //Foreign Key from Contact -> id
private int contactId;
//Constructors, Getters and Setters
}Hibernate生成的查询如下所示：
select subscriber0_.id as id1_, subscriber0_.contact_id as contact2_1_ from subscriber subscriber0_我无法从contacts表中获取联系人详细信息。我怎么才能做到这一点呢？
回答 1
Stack Overflow用户
回答已采纳
发布于 2012-04-14 03:43:15
尝试使用以下命令：
<many-to-one name="contact"
class="com.DBNAME.model.Contact" column="contact_id"
unique="true" lazy="false"/>即lazy="false"和无fetch属性。
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原文链接：
https://stackoverflow.com/questions/10147294
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