RE: [Xdoclet-user] xdoclet - @hibernate-subclass howto

2003-08-15 Thread Rupp, Heiko
You need to add discriminators: /** * @hibernate.class table="Asset" * discriminator-value="Ass" * @hibernate.discriminator column="subclass" */ public abstract class Asset { .. } and then in the subclass: /** * @hibernate.subclass * name="RegisterBoundGood" * discriminator-val

[Xdoclet-user] xdoclet - @hibernate-subclass howto

2003-08-14 Thread Russell Simpkins
How do you use xdoclet and for hibernate with a subclass? If i have class b that extends a, a and b have the same id. What do i have to put for xdoclet tags in a and in b i have in class a @hibernate-class table="a" in class b i have @hibernate-subclass the docs do not suggest more of a