对含有many-to-one外键的查询的问题

[ming1683]

<%
org.hibernate.Session s = HibernateSessionFactory.currentSession();
String hql = "from Xmb where dl=:dl order by xh asc";
Lbb dl = (Lbb)s.get(Lbb.class, new Integer(1));
List list = s.createQuery(hql).setEntity("dl", dl).list();
Xmb xmb = null;
for (Iterator its = list.iterator(); its.hasNext(); ) {
xmb = (Xmb)(Object)its.next();
。。。。。（读数据）
}
HibernateSessionFactory.closeSession();
xmb = null;
list = null;
hql = null;
%>


报错：
org.hibernate.exception.SQLGrammarException: could not execute query
=========================

Lbb dl = (Lbb)s.get(Lbb.class, new Integer(1));是没有问题的，试验过，成功得到持久化类
不知出了什么问题？

[redex]

http://community.csdn.net/Expert/topic/4970/4970357.xml?temp=5.991763E-02

[ming1683]

是List list = s.createQuery(hql).setEntity("dl", dl).list();的list()这里报错

[ming1683]

把hibernate配置文件中的show_sql开关打开，看控制台打印的sql语句是什么
——不好意思，怎么看控制台？服务在后台，没有开控制台窗口呀？

[ming1683]

知道怎么做了。
控制台输出的SQL语句信息如下：
Hibernate: select xmb0_.xh as xh3_, xmb0_.xmm as xmm3_, xmb0_.lbh as lbh3_, xmb0_.dlh as dlh3_, xmb0_.lx as lx3_, xmb0_.yxhj as yxhj3_, xmb0_.yy as yy3_, xmb0_.dx as dx3_, xmb0_.pj as pj3_, xmb0_.djsj as djsj3_, xmb0_.xgsj as xgsj3_, xmb0_.author as author3_, xmb0_.authorurl as authorurl3_, xmb0_.tp as tp3_, xmb0_.passwd as passwd3_, xmb0_.jj as jj3_, xmb0_.lr as lr3_, xmb0_.tj as tj3_, xmb0_.sc as sc3_ from softdown.xmb xmb0_ where dl=? order by xmb0_.xh asc
看上去没什么不对的呀？

[ming1683]

解决了。取了别名就正常了：

from Xmb x where x.dl = :dl order by x.xh asc