Grails2.4中JPA@MappedSuperclass的实现

Grails2.4中JPA@MappedSuperclass的实现,jpa,gorm,grails-2.4,Jpa,Gorm,Grails 2.4,我找到了解决办法,但解决不了问题。在JPA中,我们可以这样做: @MappedSuperclass public class BasicEntity { @Id @GeneratedValue(strategy = GenerationType.AUTO) private Long id; @Column(updatable = false) @Temporal(TemporalType.TIMESTAMP) private Date crea

我找到了解决办法,但解决不了问题。在JPA中,我们可以这样做:

@MappedSuperclass
public class BasicEntity {

    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    private Long id;

    @Column(updatable = false)
    @Temporal(TemporalType.TIMESTAMP)
    private Date created = new Date();

    @Temporal(TemporalType.TIMESTAMP)
    private Date modified = new Date();
}

@Entity
public class User extends BasicEntity {

    private String username;
    private String password;
}
然后,hibernate.hbm2ddl.auto生成一个包含所有继承列的表,这正是我想要的:

user (
    id,
    created,
    modified,

    username,
    password,
)
在Grails中,我这样做

abstract class BasicEntity {

    static mapping = {
        tablePerSubclass true
    }

    Date dateCreated
    Date lastUpdated
}

class User extends BasicEntity {

    String username
    String password
}
它生成了两个没有继承的表

basic_entity (
    id,
    version,
    date_created,
    last_updated,
)

user (
    id,
    username,
    password,
)
可能重复的