原因： 有一个具有相同值的不同对象已经与和 session 相关联。

解决（程序的一部分）：
 //CusClientVO cusVo = new CusClientVO();
        CusClientVO cusVo = null;
        IUserDetails user = CurrentUser.getUser();
        try {
            if ("Same".equals(cusFlag)) {//被保险人和投保人是同一人
                cusVo = (CusClientVO)this.getById(CusClientVO.class, applicant.getCAppCde());
                if(cusVo==null){  //表示新增客户。
                    cusVo = new CusClientVO();
                }
                this.saveCusApplicant(applicant, cusVo);
                if(!applicantflag ){
                    cusVo.setTCrtTm(new Date());
                    cusVo.setCCrtCde(user.getOpRelCde());
                }
            }
            cusVo.setTUpdTm(new Date());
            cusVo.setCUpdCde(user.getOpRelCde());
            //
            this.saveOrUpdate(cusVo);
        } catch (DaoException ex) {
            BusinessServiceException e = new BusinessServiceException(ex);
            e.setErrorMsg("保存客户信息失败");
        }

相关连接：

http://hi.baidu.com/anypcao/blog/item/0db1a41cd3478d8187d6b6d2.html

http://agreal.javaeye.com/blog/339560

在answer.hbm.xml中有：
 <many-to-one name="question" class="org.lxh.myzngt.vo.Question" fetch="select">
            <column name="qid" />
 </many-to-one>

所以SQL语句如下：

public List queryByUserAnswer(String userid, int currentPage, int lineSize) {
        List all = null;
        String hql = "from Question as q where q.qid in(select a.question.qid from Answer as a where a.userid=?)";
        Query q = super.getSession().createQuery(hql);
        q.setString(0, userid);
        // 分页操作。
        q.setFirstResult((currentPage - 1) * lineSize);
        q.setMaxResults(lineSize);        all = q.list();
        return all;
    }

否则报错：

org.hibernate.QueryException: could not resolve property: qid of: org.lxh.myzngt.vo.Answer [select count(q.qid) from org.lxh.myzngt.vo.Question as q where q.qid in(select a.qid from org.lxh.myzngt.vo.
Answer as a where a.userid=?)]

posted @ 2009-02-23 15:58 断点 阅读(173) | 评论 (0)

出现这个问题，主要是hibernate语句不对。

User.hbm.xml为：

<hibernate-mapping>
<class name="org.lxh.myzngt.vo.User" table="user" >
        <id name="id" type="java.lang.Integer">
            <column name="id" />
            <generator class="native"></generator>
        </id>
</class>
</hibernate-mapping>

因此，hql 语句如下：

String hql = "from User as u where u.userid=?"