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hibernate中关于HQL小疑点请问下大家

发布时间: 2012-06-05 13:54:06 作者: rapoo

hibernate中关于HQL小问题请教下大家,谢谢
写了一个测试程序,在执行到query = session.createQuery(hql); 发生异常,没报错,但跳转到HibernateTemplate不相关的类。
如果将hql改为String hql = " from User a where a.name =1";就能正常执行,为什么啊?纠结了半天,请教下大家,谢谢
分数不够了,谢谢大家
public Object doInHibernate(Session session)
throws HibernateException, SQLException
{
String hql = " from User a where a.name =:name";
Query query = null;
try
{
query = session.createQuery(hql);
}
catch (Exception e)
{
e.printStackTrace();
}

return query.list();
}
}

[解决办法]
query = session.createQuery(hql);
query.setString("name", name);
[解决办法]
String hql = " from User a where a.name =:name";这样写都有问题,你如果要传值就要用 String hql = " from User a where a.name = "+"1";或者是 String hql = " from User a where a.name ="+name;然后给name变量赋值才对啊
[解决办法]
String hql = " from User a where a.name =?";
Query query = session.createQuery(hql);
query.setString(0, name);
[解决办法]
你的name只是个键这个键没有设置值,
query.setParameter("name","name");OK

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