[英]select all from a JPQL query, with paremeters
我在平面数据上有选择参数,只知道如何完全省略参数,或者使其成为完整的通配符。 搜索可能使用一个或所有参数。 这是怎么做到的? 随着ANY
或ALL
? 或者,还有另一种方式吗?
我想对所有参数使用一个通用查询,并传递“全部”或“任何”,这些行中的某些参数。
现有代码:
package legacy.database;
import java.sql.Timestamp;
import java.util.List;
import java.util.logging.Logger;
import javax.persistence.EntityManager;
import javax.persistence.EntityManagerFactory;
import javax.persistence.Persistence;
import javax.persistence.Query;
public class MyQueries {
private static final Logger log = Logger.getLogger(MyQueries.class.getName());
private final EntityManagerFactory emf = Persistence.createEntityManagerFactory("LegacyDatabasePU");
private final EntityManager em = emf.createEntityManager();
public MyQueries() {
}
public List<Clients> findAll() {
Query q = em.createQuery("select c from Clients c");
List<Clients> clients = q.getResultList();
return clients;
}
public List<Clients> selectWithParameters(Criteria c) {
log.info(c.toString());
String opener = c.getOpener();
String closer1 = c.getCloser1();
String status = c.getStatus();
Query q = em.createQuery(
"SELECT c FROM Clients c "
+ "WHERE c.status like :status "
+ "and c.closer1 like :closer1 "
+ "and c.opener like :opener");
q.setParameter("opener", opener);
q.setParameter("closer1", closer1);
q.setParameter("status", status);
log.info(q.toString());
List<Clients> clients = q.getResultList();
log.fine(clients.toString());
return clients;
}
public Clients findById(int id) {
Clients client = em.find(Clients.class, id);
return client;
}
public void send(int id) {
Clients c = em.find(Clients.class, id);
java.util.Date date = new java.util.Date();
Timestamp t = new Timestamp(date.getTime());
em.getTransaction().begin();
c.setDateUpdated(t.toString());
em.getTransaction().commit();
}
}
如果参数是可选的,则标准API提供了一些更大的灵活性。 如果经常调用selectWithParameters ,请考虑使用参数,因为DB可以缓存参数化查询。
带有可选参数的selectWithParameters读取如下:
public List<Clients> selectWithParameters(Criteria criteria) {
log.info(criteria.toString());
String opener = criteria.getOpener();
String closer1 = criteria.getCloser1();
String status = criteria.getStatus();
CriteriaBuilder criteriaBuilder = em.getCriteriaBuilder();
CriteriaQuery<Clients> query = criteriaBuilder.createQuery(Clients.class);
Root<Clients> c = query.from(Clients.class);
List<Predicate> wherePredicates = new LinkedList<Predicate>();
if (null != status) {
wherePredicates.add(criteriaBuilder.like(c.get("status"), status));
}
if (null != closer1) {
wherePredicates.add(criteriaBuilder.like(c.get("closer1"), closer1));
}
if (null != opener) {
wherePredicates.add(criteriaBuilder.like(c.get("opener"), opener));
}
query.where(wherePredicates.toArray(new Predicate[0]));
List<Clients> clients = em.createQuery(query).getResultList();
log.fine(clients.toString());
return clients;
}
海纳,谢谢你。 这工作,不知道为什么我遇到海纳的代码有问题,但他的样本让我朝着正确的方向:
public List<Clients> selectByCriteria(Criteria criteria) {
CriteriaBuilder criteriaBuilder = em.getCriteriaBuilder();
CriteriaQuery<Clients> clientCriteriaQuery = criteriaBuilder.createQuery(Clients.class);
Root<Clients> clientRoot = clientCriteriaQuery.from(Clients.class);
clientCriteriaQuery.select(clientRoot);
List<Predicate> predicates = new ArrayList<>();
predicates.add(criteriaBuilder.like(clientRoot.get(Clients_.phone1), "%" + criteria.getPhone1() + "%"));
if (!criteria.getOpener().equalsIgnoreCase("all")) {
predicates.add(criteriaBuilder.like(clientRoot.get(Clients_.opener), "%" + criteria.getOpener() + "%"));
}
if (!criteria.getCloser1().equalsIgnoreCase("all")) {
predicates.add(criteriaBuilder.like(clientRoot.get(Clients_.closer1), "%" + criteria.getCloser1() + "%"));
}
if (!criteria.getStatus().equalsIgnoreCase("all")) {
predicates.add(criteriaBuilder.like(clientRoot.get(Clients_.status), "%" + criteria.getStatus() + "%"));
}
clientCriteriaQuery.where(predicates.toArray(new Predicate[0]));
List<Clients> clients = em.createQuery(clientCriteriaQuery).getResultList();
return clients;
}
海纳的回答可能没有实质性的差异(?)。 JPA和JPQL有点模糊。 我简直不敢相信,但我几乎更喜欢SQL! 我得调整一下。
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