ORDER BY使用Criteria API

Question

当我写一个HQL查询

Query q = session.createQuery("SELECT cat from Cat as cat ORDER BY cat.mother.kind.value");
return q.list();

一切都好.但是,当我写一个Criteria时

Criteria c = session.createCriteria(Cat.class);
c.addOrder(Order.asc("mother.kind.value"));
return c.list();

我得到一个例外 org.hibernate.QueryException: could not resolve property: kind.value of: my.sample.data.entities.Cat

如果我想使用Criteria and Order,我应该如何表达我的"order by"？

Answer 1

您需要为.创建别名mother.kind.你是这样做的.

Criteria c = session.createCriteria(Cat.class);
c.createAlias("mother.kind", "motherKind");
c.addOrder(Order.asc("motherKind.value"));
return c.list();

Answer 2

在没有看到映射的情况下很难确定(请参阅@ Juha的评论),但我认为您需要以下内容:

Criteria c = session.createCriteria(Cat.class);
Criteria c2 = c.createCriteria("mother");
Criteria c3 = c2.createCriteria("kind");
c3.addOrder(Order.asc("value"));
return c.list();

Answer 3

您还可以添加联接类型：

Criteria c2 = c.createCriteria("mother", "mother", CriteriaSpecification.LEFT_JOIN);
Criteria c3 = c2.createCriteria("kind", "kind", CriteriaSpecification.LEFT_JOIN);

Answer 4

这是你不得不做的事情,因为不推荐使用sess.createCriteria:

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();