Question

我尝试使用此映射：

@Entity
@Table(name="ecc.\"RATE\"")
@Inheritance(strategy=InheritanceType.SINGLE_TABLE)
@DiscriminatorColumn(name="DISCRIMINATOR", discriminatorType= DiscriminatorType.STRING)
public abstract class Rate extends GenericBusinessObject {
...
}

@Entity
@DiscriminatorValue("E")
public class EntranceRate extends Rate { 
 @ManyToOne
 @JoinColumn(name = "\"RATES_GRID_ID\"")
 protected RatesGrid ratesGrid;
...
}


@Entity
@Table(name="ecc.\"RATES_GRID\"")
public class RatesGrid extends GenericBusinessObject {
 /** */
 @OneToMany(mappedBy = "ratesGrid",  targetEntity = EntranceRate.class, fetch=FetchType.LAZY)
 private List<EntranceRate> entranceRates;
}

当我尝试从entranceRates对象访问我的ratesGrid列表时，出现此错误：

Object with id: 151 was not of the specified subclass: com.ecc.bo.rate.EntranceRate (loaded object was of wrong class class com.ecc.bo.rate.AnnualRate)

查看生成的sql，我在where子句中找不到“discriminator =”的痕迹。我做错了什么？

我使用PostGreSQL数据库和Hibernate作为JPA提供程序。

Answer 1

我不知道这是一个错误还是一个功能（对我而言，这是一个错误），但解决方案（解决方法？）是在顶级类上使用Hibernate注释@ForceDiscriminator：

@Entity
@Table(name="ecc.\"RATE\"")
@Inheritance(strategy=InheritanceType.SINGLE_TABLE)
@DiscriminatorColumn(name="DISCRIMINATOR", discriminatorType= DiscriminatorType.STRING)
@org.hibernate.annotations.ForceDiscriminator
public abstract class Rate extends GenericBusinessObject {
    ...
}

您可能想投票给HHH-4358。

JPA：继承 - 在生成的SQL中未考虑Discriminator值

1 个答案: