hibernate 注解如何绑定自定义oracle sequence

[chengerxiang]

数据库表手动建立，grade表，字段id,name,student表，字段id,name,gradeid(外键，引用grade表主键），为grade建立序列seq_grade_id,为student表建立序列seq_student_id,两个序列都是从0开始，每次增1，最小值0，类中，在Student类中保存Grade类的引用，单向多对一的关系，该如何使用注解配置

[computerclass]

http://blog.sina.com.cn/s/blog_4b5bc01101010p4x.html

[lina791211]

引用 1 楼 Chen_yanxia 的回复:

<class name="com.prj.bean.Grade" table="Grade" schema="SCOTT"> <id name="id" type="java.lang.Integer"> <column name="ID" precision="22" scale="0" /> <generator class="sequence"> <param name="sequence">seq_grade_id</param> </generator> </id>

2L正解 应该不用关注大小写（我自己全是大写处理）

[lina791211]

<hibernate-mapping>
    <class name="org.ds.vo.Userinfo" table="USERINFO" schema="HRDC">
        <id name="userid" type="java.lang.Long">
            <column name="USERID" precision="22" scale="0" />
            <generator class="sequence" >
            	<param name="sequence">seq_userinfo</param>
            </generator>
        </id>
        <property name="username" type="java.lang.String">
            <column name="USERNAME" length="50" />
        </property>
        <property name="password" type="java.lang.String">
            <column name="PASSWORD" length="50" />
        </property>
</hibernate-mapping>

[Fate-]

<class name="com.prj.bean.Grade" table="Grade" schema="SCOTT"> <id name="id" type="java.lang.Integer"> <column name="ID" precision="22" scale="0" /> <generator class="sequence"> <param name="sequence">seq_grade_id</param> </generator> </id>

[launchit]

直接用native 就ok

[i9577]

话说人家说的是注解啊 @Id @GeneratedValue(generator = "paymentableGenerator") @GenericGenerator(name = "paymentableGenerator", strategy = "sequence", parameters = { @Parameter(name = "sequence", value = "sequence_types") }) @Column(name = "ID", unique = true, nullable = false, precision = 22, scale = 0)