如何使用hibernate.list

<list
            name="roles"
            lazy="true"
            inverse="true"
            cascade="none"
        >

              <key
                  column="realm_id"
              />

              <one-to-many
                  class="com.entity.auth.Role"
              />
        </list>

错的么？list里面不允许有one-to-many的关系？

list需要<index column="XXX"/>

谢谢yesh220

再问一下
用ant下面代码：

        <java classname="net.sf.hibernate.tool.hbm2ddl.SchemaExport" fork="true">
            <classpath refid="hibernate.ext.classpath"/>
           
            <arg value="--text"/>
            <arg value="--output=db/schema.sql"/>
            <arg value="--properties=src/hibernate.properties"/>
            <arg value="--delimiter=;"/>
            <arg value="src/java/entity/auth/*.xml"/>
        </java>

需要depends java代码编译么？

另外： * @hibernate.list role  = "roles-accounts"
*                 table = "roles_accounts"
*                 lazy  = "true"
* @hibernate.collection-index
*                 index = "seq"
*                 type  = "long"
*                  
* @hibernate.collection-key
*                 column = "role_id"
* @hibernate.collection-many-to-many
*                 column = "account_id"
*                 class = "com.persistence.entity.Account"

其中的index="XXX",XXX是任意值？

查看文档中collection那部分