java - JPA CriteriaBuilder 获取具有条件的列表实体

Question

如何使用 JPA CriteriaBuilder 进行此选择，就像我的示例中的 hibernate Criteria 一样？

Criteria criteria = getSession().createCriteria(Payment.class);
        criteria.setFirstResult(0);
        criteria.setMaxResults(1);
        criteria.add(Restrictions.eq("statusStepOne", 0));
        List list = criteria.list();

和

Criteria criteria = getSession().createCriteria(Payment.class);
        criteria.add(Restrictions.eq("oid", oid));
        criteria.add(Restrictions.between("dateCreated", Utils.getDateMinus(1), new Date()));
        List list = criteria.list();
        return (List<Payment>)list;

Answer 1

第一个查询

   Criteria criteria = getSession().createCriteria(Payment.class);
   criteria.setFirstResult(0);
   criteria.setMaxResults(1);
   criteria.add(Restrictions.eq("statusStepOne", 0));
   List list = criteria.list();

JPA 第一个查询

   CriteriaBuilder cb = em.getCriteriaBuilder();
   CriteriaQuery<Payment> query = cb.createQuery(Payment.class);

   Root<Payment> p = query.from(Payment.class);
   query.select(p);
   query.where(cb.equal(p.get("statusStepOne"), 0));
   query.setFirstResult(0);
   query.setMaxResults(1);

   List<Payment> list = query.getResultList();

第二个查询

   Criteria criteria = getSession().createCriteria(Payment.class);
   criteria.add(Restrictions.eq("oid", oid));
   criteria.add(Restrictions.between("dateCreated", Utils.getDateMinus(1), new Date()));
   return (List<Payment>) criteria.list(); // I've removed useless asignment

JPA 第二个查询

  CriteriaBuilder cb = em.getCriteriaBuilder();
  CriteriaQuery<Payment> query = cb.createQuery(Payment.class);

  Root<Payment> p = query.from(Payment.class);
  query.select(p);
  query.where(cb.equal(p.get("oid"), oid)
       .and(cb.between(p.get("dateCreated"), Utils.getDateMinus(1), new Date())));