休眠子选择与类继承

Question

我有相同的字段的两个对象，而是从不同的表（事实上foo_something和bar_something不是简单的表格，有几个结果连接操作）。我使用@Subselect hibernate注释来获取数据。我可以做这样的班级结构，但哪些工作正常？此类结构不起作用，因为注释不会继承。如果我使用@Inheritance注解，我得到错误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 { 

} 

UPD。

谢谢，@HenryMartens，我加了注释@Entity和@Inheritance与TABLE_PER_CLASS战略，并提出了抽象类，现在它的正常工作：

@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

在Hibernate中DTYPE是鉴别列。它用于区分一些对象，它们与您的示例一样保存在同一个数据库表中。 你可以用@DiscriminatorColumn添加注释的鉴别列。

在具体的类可以使用@DiscriminatorValue设置鉴别列的值注释类。

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 { 

}