Question

我想使用条件编写以下查询。

我需要找到distinct行，并在where子句中使用当前日期。我如何在Criteria中实现这一点。

SELECT DISTINCT *
FROM EMAIL_PROGRAM
WHERE CURRENT_DATE >=PGM_START_DT
AND CURRENT_DATE   <= PGM_END_DT
AND EMAIL_PGM_FLG     ='Y'
AND EMAIL_PGM_DESC  IS NOT NULL
and RGN_CD = 'US';

以下是我需要申请的代码。

SessionFactory factory = null;
    Session session = null;
    try {
        factory = getSessionFactory();
        session = factory.openSession();

        final Criteria criteria = session
                .createCriteria(EmailDataBean.class);
        returnList = criteria.list();
    } catch (Exception e) {
        logger.error(e.getMessage());
        throw new DAOException(e);
    } finally {
        DBUtil.close(factory, session);
    }
    if (logger.isInfoEnabled()) {
        logger.info(LOG_METHOD_EXIT);
    }
    return returnList;
}

Answer 1

您可以在标准对象上使用以下内容。

criteria.setResultTransformer(Criteria.DISTINCT_ROOT_ENTITY);

Answer 2

您应该执行类似

的操作

//first add conditions in the where clause (you should use property names as defined in your Hbernate entities , not column names in your DB)
criteria.add(Restrictions.ge("PGM_START_DT", startDt));
criteria.add(Restrictions.le("PGM_END_DT", endDt));
criteria.add(Restrictions.eq("EMAIL_PGM_FLG", "Y"));
criteria.add(Restrictions.isNotNull("EMAIL_PGM_DESC"));
criteria.add(Restrictions.eq("RGN_CD", "US"));

现在，将每个列（即Hibernate实体属性/字段）添加到Projection列表（这需要支持查询中的 distinct ）

ProjectionList pList = Projections.projectionList();
pList.add(Projections.property("PGM_START_DT"), "PGM_START_DT");
pList.add(Projections.property("PGM_END_DT"), "PGM_END_DT");
// add all the other properties (columns) and then have the Projections.distinct method act on the entire row (all the columns)
criteria.setProjection(Projections.distinct(pList));

默认情况下，使用Projections确实会将结果作为List＆lt; Object []＆gt;返回而不是List＆lt; EmailDataBean＆gt;，这通常不方便。要解决这个问题，您应该设置ResultTransformer

crit.setResultTransformer(Transformers.aliasToBean(EmailDataBean.class));

或者，您可以使用

代替使用投影

criteria.setResultTransformer(Criteria.DISTINCT_ROOT_ENTITY);

但是这不会从数据库中获取不同的行，而是让Hibernate过滤结果（删除重复项）。

在休眠标准中写出明显的

2 个答案: