我相信我有这个工作,我一定做了些什么。希望有人可以立即看到问题。
继承关系中的两个类。两张桌子在玩。两个表都接收记录。手头的问题:没有存储鉴别器值。
父类
@Entity
@Inheritance(strategy=InheritanceType.JOINED)
@DiscriminatorColumn(name="entp", discriminatorType=DiscriminatorType.STRING)
@Table(name="enrg")
public abstract class BaseEntity implements Serializable {
private static final long serialVersionUID = 1L;
@Id
@GeneratedValue(strategy=GenerationType.AUTO)
@Column(name="enid")
private Long entityId;
// some string property mapped to column 'nm;
// some string property mapped to column 'ns;
// some string property mapped to column 'tmcr;
儿童班
@Entity
@DiscriminatorValue("ws")
@Table(name="ws")
public class Website extends BaseEntity {
private static final long serialVersionUID = 1L;
@Column(name="exdr", length=100)
private String exportDirectory;
服务电话
getEntityManager().persist(website);
SQL
Hibernate: insert into enrg (nm, ns, tmcr) values (?, ?, ?)
Hibernate: insert into ws (exdr, enid) values (?, ?)
休眠
<persistence-unit name="prod" transaction-type="JTA">
<provider>org.hibernate.ejb.HibernatePersistence</provider>
<jta-data-source>prodDataSource</jta-data-source>
<class>webadmin.domain.core.Website</class>
<exclude-unlisted-classes>false</exclude-unlisted-classes>
<properties>
<property name="hibernate.dialect" value="org.hibernate.dialect.MySQL5Dialect" />
<property name="hibernate.connection.driver_class" value="com.mysql.jdbc.Driver" />
<property name = "hibernate.show_sql" value = "true" />
<property name = "hibernate.discriminator.ignore_explicit_for_joined" value = "false" />
</properties>
</persistence-unit>
我需要做什么,让Hibernate在enrg的插入中包含entp列/值?在这种情况下,entp值应为&#39; ws&#39;
解决方案
我的所有映射/配置仍然如上所述。我已升级到Hibernate 4.3.6,在TomEE 1.7下,现在可以使用了。
Hibernate: insert into enrg (nm, ns, tmcr, entp) values (?, ?, ?, 'ws')
Hibernate: insert into ws (exdr, enid) values (?, ?)