Java EclipseLink：从基本实体中选择，并按InheritanceType.JOINED的子字段进行筛选

我有以下设置

@Entity
@Table(name = "FOO")
@Inheritance(strategy = InheritanceType.JOINED)
@DiscriminatorColumn(name = "FOO_TYPE", discriminatorType = DiscriminatorType.STRING,     length = 20)
@DiscriminatorValue("FOO")
public class Foo implements Serializable {

  private static final long serialVersionUID = 1L;
  @Id
  @Column(name = "ID")
  @GeneratedValue(strategy = GenerationType.IDENTITY)
  private Long id;

  @Column(name = "foo_column")
  private String fooColumn;

  ... getters and setters ...

}


@Entity
@Table(name = "BAR")
@PrimaryKeyJoinColumn(name = "BAR_ID", referencedColumnName = "ID")
@DiscriminatorValue("Bar")
public class Bar extends Foo {

  @Column(name = "bar_column")
  private String barColumn;

  ... getters and setters ...

}

当我以JPA提供者的身份在Hibernate 3.6中执行以下查询时，一切都按预期进行（至少……正如我预期的那样），结果列表中填充了扩展Foo类的各种对象：

List<Foo> fooList = em.createQuery("SELECT foo FROM Foo foo WHERE 1=1 AND foo.fooColumn = 'foo' and foo.barColumn = 'bar'");

注意：我已经阅读了EclipseLink的文档，并专门在表中添加了@DiscriminatorColumn和@DiscriminatorValue

然后，我使用TREAT函数将查询更改为以下内容：

List<Foo> fooList = em.createQuery("SELECT foo FROM Foo foo WHERE 1=1 AND foo.fooColumn = 'foo' and treat(foo as Bar).barColumn = 'bar'");

List-doulist=em.createQuery（“从foo-foo中选择foo，其中1=1，foo.fooColumn='foo'并处理（foo为Bar.barColumn='Bar'）；

现在可以使用了（根据JPA2.1）

---

是否可以在不借助JPA 2.1功能的情况下对子实体进行过滤？

是的，您只需查询子实体而不是父实体。您可以使用TREAT功能进行测试吗？类似于：

SELECT b.name，b.ISBN FROM Order o JOIN TREAT（o.product AS Book）b

List<Foo> fooList = em.createQuery("SELECT foo FROM Foo foo WHERE 1=1 AND foo.fooColumn = 'foo' and treat(foo as Bar).barColumn = 'bar'");