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A 类有A1,A2两个subclass .
B1类一对多个A1 ,
B2类一对过个A2 .
本想，通过B1实例获得A1 Set , 但现在通过B1只能获得的是A set .
问，如何配置B1,以便获得A1 Set ?
具体的配置文件
A.bhm
<discriminator column="flag" type="string" />
<subclass name="A1" discriminator-value="0" >
<many-to-one
name="account"
column="account_id"
class="B1"
not-null="false"
>
</many-to-one>
</subclass>
<subclass name="A2" discriminator-value="0" >
<many-to-one
name="account"
column="account_id"
class="B2"
not-null="false"
>
</many-to-one>
</subclass>
==========================================================
B1.bhm
<set name="As" inverse="true" lazy="true">
<key column="account_id"/>
<one-to-many class="A1"/>
</set>
up
我也遇到这样的问题呢！
<hibernate-mapping
>
<class
name="B1
table="table_b1"
dynamic-update="true"
dynamic-insert="true"
optimistic-lock="version"
>
<id
name="id"
column="account_id"
type="java.lang.Integer"
>
<generator class="sequence">
<param name="sequence">oracle_sequence</param>
</generator>
</id>
<set name="weighs"
lazy="true"
inverse="true"
cascade="none"
>
<key column="account_id">
</key>
<one-to-many class="A1"/>
</set>
</class>
</hibernate-mapping>
====================================
貌似你的A.xml里的many-to-one漏了个class的属性
应该是可以地
<subclass name="A1" discriminator-value="0" >
<subclass name="A2" discriminator-value="0" >
两句有问题，感兴趣的话来QQ群15292374来聊啊，具体解决
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