Question

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

Answer 1

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

Answer 2

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
 * 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<Z,X> extends Join<Z,X>, Fetch<Z,X>, FromImplementor<Z,X> {
    /**
     * {@inheritDoc}
     * <p/>
     * Refined return type
     */
    public JoinImplementor<Z,X> correlateTo(CriteriaSubqueryImpl subquery);
}

Answer 3

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));

Answer 4

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。

Answer 5

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);

Answer 6

丑陋但是：

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

JPA 2条件获取路径导航

6 个答案: