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您好我是JPA Criteria生成器概念的新手,我使用PostgreSQL和JPA Query.I獲得此查詢SELECT id,full_name,email FROM nurses WHERE(lower(sender)LIKE' %bar%'和lower(receiver)LIKE'%bar%'),但是如何將其轉換爲JPA Criteria構建器。如何使用JPA查詢使大小寫不敏感PostgreSQL
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();