具有类继承的休眠子选择

Question

我有两个具有相同字段的对象，但来自不同的表(foo_something和bar_something实际上不是普通表，有几次连接操作的结果)。我使用休眠注释@Subselect以获取数据。我可以像这样制作类结构吗，但是哪个工作正常？此类结构不起作用，因为批注不会继承。如果我使用@Inheritance注释，则会出现错误Unknown column foo_somethingca0.DTYPE

class Something {
@Id @Column Long id;
@Column String name;
@Column String description;
//getters, setters, constructors
}
@Entity
@Immutable
@Subselect("SELECT id, name, description FROM foo_something")
class FooSomething extends Something {
}
@Entity
@Immutable
@Subselect("SELECT id, name, description FROM bar_something")
class BarSomething extends Something {
}

上。

谢谢，@HenryMartens，我添加了注释@Entity和@InheritanceTABLE_PER_CLASS策略，并使类抽象，现在它工作正常：

@Entity
@Inheritance(strategy = TABLE_PER_CLASS)
abstract class Something {
@Id @Column Long id;
@Column String name;
@Column String description;
//getters, setters, constructors
}
@Entity
@Immutable
@Subselect("SELECT id, name, description FROM foo_something")
class FooSomething extends Something {
}
@Entity
@Immutable
@Subselect("SELECT id, name, description FROM bar_something")
class BarSomething extends Something {
}

Answer 1

Hibernate中的DTYPE是鉴别器列。它用于区分保留在同一数据库表中的某些对象，如您的示例所示。 您可以添加带有@DiscriminatorColumn注释的鉴别器列。

在具体类上，您可以使用@DiscriminatorValue为带注释的类设置鉴别器列的值。

例如：

@Entity
@Inheritance( strategy = InheritanceType.TABLE_PER_CLASS ) 
@DiscriminatorColumn(name="DISC", discriminatorType=STRING, length=20)
class abstract Something {
@Id 
@Column 
Long id;
@Column 
String name;
@Column 
String description;
//getters, setters, constructors 
}
@Entity
@Table("FOO_SOMETHING")
@Immutable
@DiscriminatorValue("FooSomething")
class FooSomething extends Something {
}
@Entity
@Table("BAR_SOMETHING")
@Immutable
@DiscriminatorValue("BarSomething")
class BarSomething extends Something {
}