【发布时间】:2017-09-11 14:47:44
【问题描述】:
您好,我是 JPA Criteria builder 概念的新手,我正在使用 PostgreSQL 和 JPA Query。我得到了这个查询 SELECT id, full_name, email FROM Nurses WHERE(lower(sender) LIKE '%bar%' and lower(receiver) LIKE '%bar%')。但是如何将其转换为 JPA Criteria builder。
entityManager.getTransaction().begin();
CriteriaBuilder cb = entityManager.getCriteriaBuilder();
System.out.println(entityManager + " Conected");
CriteriaQuery<sourceTracking> cq = cb.createQuery(sourceTracking.class);
Root<sourceTracking> data1 = cq.from(sourceTracking.class);
cq.where(cb.like(data1.<String>get("sender"), "%"+sender+"%"),cb.like(data1.<String>get("receiver"), "%"+receiver+"%");
cq.select(data1);
TypedQuery<sourceTracking> tquery = entityManager.createQuery(cq);
sourselList = tquery.getResultList();
【问题讨论】:
标签: java postgresql hibernate jpa