getHibernateTemplate()用法

Question

public List findByDeviceIdsRasId(final String rasId, final List deviceIds)
throws DataAccessException
{
if (StringUtils.isEmpty(rasId) || CollectionUtils.isEmpty(deviceIds))
{
return null;
}
/*return getHibernateTemplate().execute(new HibernateCallback>()
{

        @Override
        public List<RasPo> doInHibernate(Session session) throws HibernateException,
                SQLException
        {
            StringBuffer stringBuffer = new StringBuffer();
            stringBuffer.append("(");
            for (String deviceId : deviceIds)
            {
                stringBuffer.append("'").append(deviceId).append("'").append(",");
            }
            String str = stringBuffer.substring(0, stringBuffer.length() - 1);
            str = str + ")";
            Query query= session.createQuery("from RasPo where rasId='" + rasId + "' and deviceId in " + str);
            List<RasPo> list = query.list();
            return list;
        }
    });*/
    List<RasPo> list=null;
    try{
        StringBuffer stringBuffer = new StringBuffer();
        stringBuffer.append("(");
        for (String deviceId : deviceIds)
        {
            stringBuffer.append("'").append(deviceId).append("'").append(",");
        }
        String str = stringBuffer.substring(0, stringBuffer.length() - 1);
        str = str + ")";
        list = getHibernateTemplate().find("from RasPo where rasId=? and deviceId in ?",
                new String[] {rasId, str});
    }
    catch (DataAccessException e)
    {
        throw e;
    }
    return list;
}

使用上面的注释的可以查询出来，但是改为find（）方法后就查询不出来了。我感觉着没有什么区别了。求解

Answer 1

ist = getHibernateTemplate().find("from RasPo where rasId=? and deviceId in ?",
new String[] {rasId, str});

第二个?是 in 不能传 一个字符串，解决方案，类似于下边：

List result = getHibernateTemplate().executeFind(new HibernateCallback>() {
@Override
public List doInHibernate(Session session) throws HibernateException, SQLException {

            String hql = "from SampleModel where uuid in (:uuids)";

            Query q = session.createQuery(hql);
            q.setParameterList("uuids", new Object[]{0, 1}); 
            return q.list();
        }

    });

1、hql使用 String hql = "from SampleModel where uuid in (:uuids)"; 在hibernate必须使用命名参数；
2、赋值 q.setParameterList("uuids", new Object[]{0, 1}); 可以是集合/数组；

hibernate会动态生成如下语句（in里边多个？）
Hibernate: select samplemode0_.uuid as uuid0_, samplemode0_.name as name0_ from tbl_sample samplemode0_ where samplemode0_.uuid in (? , ?)

如下是不支持的
list = getHibernateTemplate().find("from RasPo where rasId=? and deviceId in ?",

new String[] {rasId, str});

你的也应该改成类似于这样
hql = "from RasPo where rasId=:rasId and deviceId in (:deviceIds)";

设置参数
不能传str 而是一个deviceIds集合