相关疑难解决方法(0)

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 …

使用JPA 2.0.似乎默认情况下(没有显式提取),@OneToOne(fetch = FetchType.EAGER)在1 + N个查询中提取字段,其中N是包含定义与不同相关实体的关系的实体的结果数.使用Criteria API,我可能会尝试避免如下:

CriteriaBuilder builder = entityManager.getCriteriaBuilder();
CriteriaQuery<MyEntity> query = builder.createQuery(MyEntity.class);
Root<MyEntity> root = query.from(MyEntity.class);
Join<MyEntity, RelatedEntity> join = root.join("relatedEntity");
root.fetch("relatedEntity");
query.select(root).where(builder.equals(join.get("id"), 3));

理想情况下,上述内容应与以下内容相同:

SELECT m FROM MyEntity m JOIN FETCH myEntity.relatedEntity r WHERE r.id = 3

但是,条件查询导致根表不必要地连接到相关实体表两次; 一次用于获取,一次用于where谓词.生成的SQL看起来像这样:

SELECT myentity.id, myentity.attribute, relatedentity2.id, relatedentity2.attribute 
FROM my_entity myentity 
INNER JOIN related_entity relatedentity1 ON myentity.related_id = relatedentity1.id 
INNER JOIN related_entity relatedentity2 ON myentity.related_id = relatedentity2.id 
WHERE relatedentity1.id = 3

唉,如果我只进行获取,那么我没有在where子句中使用的表达式.

我错过了什么,或者这是Criteria API的限制？如果是后者,这是否在JPA 2.1中得到纠正,还是有任何特定于供应商的增强功能？

否则,最好放弃编译时类型检查(我意识到我的示例不使用元模型)并使用动态JPQL TypedQueries.