我有类别的列表。我需要排除2,3行类的列表。我们可以通过使用条件和限制通过Hibernate实现?

有帮助吗?

解决方案

您的问题是有些不清楚。假设“类别”是一个根实体和“2,3”是ID(或类别的某些属性的值“),则可以使用以下排除它们:

Criteria criteria = ...; // obtain criteria from somewhere, like session.createCriteria() 
criteria.add(
  Restrictions.not(
     // replace "id" below with property name, depending on what you're filtering against
    Restrictions.in("id", new long[] {2, 3})
  )
);

同样可以与DetachedCriteria来完成。

其他提示

 Session session=(Session) getEntityManager().getDelegate();
        Criteria criteria=session.createCriteria(RoomMaster.class);
//restriction used or inner restriction ...
        criteria.add(Restrictions.not(Restrictions.in("roomNumber",new String[] { "GA8", "GA7"})));
        criteria.setResultTransformer(Criteria.DISTINCT_ROOT_ENTITY);
        List<RoomMaster> roomMasters=criteria.list();

有关,因为新的标准版本5.2的Hibernate:

CriteriaBuilder criteriaBuilder = getSession().getCriteriaBuilder();
CriteriaQuery<Comment> criteriaQuery = criteriaBuilder.createQuery(Comment.class);

List<Long> ids = new ArrayList<>();
ids.add(2L);
ids.add(3L);

Root<Comment> root = getRoot(criteriaQuery);
Path<Object> fieldId = root.get("id");
Predicate in = fieldId.in(ids);
Predicate predicate = criteriaBuilder.not(in);

criteriaQuery
        .select(root)
        .where(predicate);

List<Comment> list = getSession()
        .createQuery(criteriaQuery)
        .getResultList();
许可以下: CC-BY-SA归因
不隶属于 StackOverflow
scroll top