JPA 2条件获取路径导航
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 c = cb.createQuery(Team.class); Root t = c.from(Team.class); Join p = t.join(Team_.players); c.select(t).where(cb.equal(p.get(Player_.age), age)); TypedQuery q = entityManager.createQuery(c); List 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 cq = cb.createQuery(Team.class); Root t = cq.from(Team.class); Fetch 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 q = entityManager.createQuery(cq); List result = q.getResultList();
根据Hiberante Documentation fetch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
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 p = t.join(Team_.players); t.fetch(Team_.players); c.select(t).where(cb.equal(p.get(Player_.age), age));
即，执行join() ，为它添加一个fetch() ，然后使用join。 这是不合逻辑的，只会增加JPA标准的不雅性，但无论如何，应该是一个变通方法
它适用于我使用Hibernate Provider。
//Join Example (default inner join) int age = 25; CriteriaBuilder cb = entityManager.getCriteriaBuilder(); CriteriaQuery c = cb.createQuery(Team.class); Root t = c.from(Team.class); // Join p = t.join(Team_.players); Join p = (Join)t.fetch(Team_.players); c.select(t).where(cb.equal(p.get(Player_.age), age)); TypedQuery q = entityManager.createQuery(c); List 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function。
*这很奇怪，因为hibernate文档没有显示这个例子。
掌握它看看这个界面。
/* * 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 * Boston, MA 02110-1301 USA */ 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 extends Join, Fetch, FromImplementor { /** * {@inheritDoc} * * Refined return type */ public JoinImplementor correlateTo(CriteriaSubqueryImpl subquery); }
您所要做的就是以下内容：
1-取得。 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 cq = cb.createQuery(Team.class); Root t = cq.from(Team.class); Fetch p = t.fetch(Team_.players); cq.where(cb.equal(t.get("player").get("age"), age));
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（）。 从问题中的示例：
//Join Example (default inner join) int age = 25; CriteriaBuilder cb = entityManager.getCriteriaBuilder(); CriteriaQuery c = cb.createQuery(Team.class); Root t = c.from(Team.class); Join p = t.join(Team_.players); c.select(t).where(cb.equal(p.get(Player_.age), age)); TypedQuery q = entityManager.createQuery(c); List result = q.getResultList();
如果这：
TypedQuery q = entityManager.createQuery(c);
被替换为：
EntityGraph fetchGraph = getEntityManager().createEntityGraph(Team.class); fetchGraph.addSubgraph(Team_.players); TypedQuery q = entityManager.createQuery(c).setHint("javax.persistence.loadgraph", fetchGraph);
然后所有球员都会渴望获得。
我正在使用JPA 2.1与Hibernate 4.3.7，以下对我很有用。 它甚至看起来都不那么难看。
Join p = (Join) t.fetch(Team_.players);
丑陋但是：
Join 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hack，它运行JBoss6.1 hibernate