当我编写HQL查询时
Query q = session.createQuery("SELECT cat from Cat as cat ORDER BY cat.mother.kind.value");
return q.list();
百事大吉。但是,当我编写一个标准时,
Criteria c = session.createCriteria(Cat.class);
c.addOrder(Order.asc("mother.kind.value"));
return c.list();
我得到一个异常org.hibernate.QueryException: could not resolve property: kind.value of: my.sample.data.entities.Cat
如果我想使用条件和顺序,我应该如何表达我的“Order by"?
发布于 2009-11-23 06:39:13
您需要为mother.kind
创建一个别名。你可以这样做。
Criteria c = session.createCriteria(Cat.class);
c.createAlias("mother.kind", "motherKind");
c.addOrder(Order.asc("motherKind.value"));
return c.list();
发布于 2009-11-23 06:02:02
如果不查看映射(请参阅@Juha的评论),就很难确定,但我认为您需要类似于以下内容:
Criteria c = session.createCriteria(Cat.class);
Criteria c2 = c.createCriteria("mother");
Criteria c3 = c2.createCriteria("kind");
c3.addOrder(Order.asc("value"));
return c.list();
发布于 2017-04-15 15:00:21
这是您必须做的,因为sess.createCriteria已被弃用:
CriteriaBuilder builder = getSession().getCriteriaBuilder();
CriteriaQuery<User> q = builder.createQuery(User.class);
Root<User> usr = q.from(User.class);
ParameterExpression<String> p = builder.parameter(String.class);
q.select(usr).where(builder.like(usr.get("name"),p))
.orderBy(builder.asc(usr.get("name")));
TypedQuery<User> query = getSession().createQuery(q);
query.setParameter(p, "%" + Main.filterName + "%");
List<User> list = query.getResultList();
https://stackoverflow.com/questions/1780129
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