使用JPA 2 Criteria Join方法,我可以执行以下操作:
//Join Example (default inner join)
int age = 25;
CriteriaBuilder cb = entityManager.getCriteriaBuilder();
CriteriaQuery<Team> c = cb.createQuery(Team.class);
Root<Team> t = c.from(Team.class);
Join<Team, Player> p = t.join(Team_.players);
c.select(t).where(cb.equal(p.get(Player_.age), age));
TypedQuery<Team> q = entityManager.createQuery(c);
List<Team> result = q.getResultList();
我怎么能用fetch方法做同样的事情,我期望Fetch接口有get路径导航方法,但它没有:
//Fetch Join Example
int age = 25;
CriteriaBuilder cb = entityManager.getCriteriaBuilder();
CriteriaQuery<Team> cq = cb.createQuery(Team.class);
Root<Team> t = cq.from(Team.class);
Fetch<Team,Player> p = t.fetch(Team_.players);
cq.where(cb.equal(p.get(Player_.age), age)); //This leads to compilation error there is no such method get in interface Fetch
TypedQuery<Team> q = entityManager.createQuery(cq);
List<Team> result = q.getResultList();
根据Hiberante文档提取返回一个错误的Join对象。 http://docs.jboss.org/hibernate/stable/entitymanager/reference/en/html/querycriteria.html#querycriteria-from-fetch
答案 0 :(得分:19)
同意你对这种方法的看法,以及你希望它允许你所说的这一事实。另一种选择是
Join<Team, Player> p = t.join(Team_.players);
t.fetch(Team_.players);
c.select(t).where(cb.equal(p.get(Player_.age), age));
即执行join()
,为其添加fetch()
,然后使用联接。这是不合逻辑的,只会增加JPA标准的不雅性,但无论如何,应该是一种解决方法
答案 1 :(得分:10)
它适用于我使用Hibernate Provider。
//Join Example (default inner join)
int age = 25;
CriteriaBuilder cb = entityManager.getCriteriaBuilder();
CriteriaQuery<Team> c = cb.createQuery(Team.class);
Root<Team> t = c.from(Team.class);
// Join<Team, Player> p = t.join(Team_.players);
Join<Team, Player> p = (Join<Team, Player>)t.fetch(Team_.players);
c.select(t).where(cb.equal(p.get(Player_.age), age));
TypedQuery<Team> q = entityManager.createQuery(c);
List<Team> result = q.getResultList();
当然,它可能会破坏可移植性,但在我们的例子中,我们一直在使用其他hibernate的独有功能。
*这很奇怪,因为hibernate文档没有显示这个例子。
要掌握它,请查看此界面。
/*
* Hibernate, Relational Persistence for Idiomatic Java
*
* Copyright (c) 2010, Red Hat Inc. or third-party contributors as
* indicated by the @author tags or express copyright attribution
* statements applied by the authors. All third-party contributions are
* distributed under license by Red Hat Inc.
*
* This copyrighted material is made available to anyone wishing to use, modify,
* copy, or redistribute it subject to the terms and conditions of the GNU
* Lesser General Public License, as published by the Free Software Foundation.
*
* This program is distributed in the hope that it will be useful,
* but WITHOUT ANY WARRANTY; without even the implied warranty of MERCHANTABILITY
* or FITNESS FOR A PARTICULAR PURPOSE. See the GNU Lesser General Public License
* for more details.
*
* You should have received a copy of the GNU Lesser General Public License
* along with this distribution; if not, write to:
* Free Software Foundation, Inc.
* 51 Franklin Street, Fifth Floor
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*/
package org.hibernate.ejb.criteria;
import javax.persistence.criteria.Fetch;
import javax.persistence.criteria.Join;
/**
* Consolidates the {@link Join} and {@link Fetch} hierarchies since that is how we implement them.
* This allows us to treat them polymorphically.
*
* @author Steve Ebersole
*/
public interface JoinImplementor<Z,X> extends Join<Z,X>, Fetch<Z,X>, FromImplementor<Z,X> {
/**
* {@inheritDoc}
* <p/>
* Refined return type
*/
public JoinImplementor<Z,X> correlateTo(CriteriaSubqueryImpl subquery);
}
答案 2 :(得分:4)
您需要做的就是:
1- Do Fetch。 2-然后,走过你想要的路径。
在你的情况下:
int age = 25;
CriteriaBuilder cb = entityManager.getCriteriaBuilder();
CriteriaQuery<Team> cq = cb.createQuery(Team.class);
Root<Team> t = cq.from(Team.class);
Fetch<Team,Player> p = t.fetch(Team_.players);
cq.where(cb.equal(t.get("player").get("age"), age));
答案 3 :(得分:3)
从JPA 2.1开始,动态实体图可用于在条件查询中获取,同时使用join()而不是fetch()。从问题中的例子:
//Join Example (default inner join)
int age = 25;
CriteriaBuilder cb = entityManager.getCriteriaBuilder();
CriteriaQuery<Team> c = cb.createQuery(Team.class);
Root<Team> t = c.from(Team.class);
Join<Team, Player> p = t.join(Team_.players);
c.select(t).where(cb.equal(p.get(Player_.age), age));
TypedQuery<Team> q = entityManager.createQuery(c);
List<Team> result = q.getResultList();
如果:
TypedQuery<Team> q = entityManager.createQuery(c);
替换为:
EntityGraph<Team> fetchGraph = getEntityManager().createEntityGraph(Team.class);
fetchGraph.addSubgraph(Team_.players);
TypedQuery<Team> q = entityManager.createQuery(c).setHint("javax.persistence.loadgraph", fetchGraph);
然后所有玩家都会渴望获得。
答案 4 :(得分:1)
我正在使用带有Hibernate 4.3.7的JPA 2.1,以下对我很有帮助。它甚至看起来都不丑。
Join<Team,Player> p = (Join) t.fetch(Team_.players);
答案 5 :(得分:0)
Join<Team, Player> p=t.fetch(Team_.players);
将在sql中使用fetch生成singel连接 但这是一个丑陋的黑客工作JBoss6.1休眠