时间戳

首页 > 解决方案 > 具有聚合内部连接的 JPA 标准查询

java - 具有聚合内部连接的 JPA 标准查询

问题描述

我正在尝试编写一个 CriteriaQuery 来查询每个城市的最新观察结果。城市由city_code 字段定义，而最新记录由observation_time 字段定义。

我可以很容易地用普通的 SQL 编写它，但我无法理解如何使用 jpa 标准 api 来完成它。

select distinct m.* from 
 (select city_code cc, max(observation_time) mo
 from observations group by city_code) mx, observations m 
 where m.city_code = mx.cc and m.observation_time = mx.mo`

标签： javahibernatejpacriteria-api

解决方案

当您为宽松的效率开放时，这是可能的。因此，首先让我们将查询转换为逻辑等价的：

select distinct m.* from observations m where 
m.observation_time = (select max(inn. observation_time) from observations inn 
                      where inn.city_code = m.city_code);

然后让我们把它翻译成 JPA CriteriaQuery：

public List<Observation> maxForEveryWithSubquery() {
    CriteriaBuilder builder = entityManager.getCriteriaBuilder();
    CriteriaQuery<Observation> query = builder.createQuery(Observation.class);
    Root<Observation> observation = query.from(Observation.class);
    query.select(observation);

    Subquery<LocalDateTime> subQuery = query.subquery(LocalDateTime.class);
    Root<Observation> observationInner = subQuery.from(Observation.class);
    subQuery.where(
            builder.equal(
                    observation.get(Observation_.cityCode),
                    observationInner.get(Observation_.cityCode)
            )
    );
    Subquery<LocalDateTime> subSelect = subQuery.select(builder.greatest(observationInner.get(Observation_.observationTime)));
    query.where(
            builder.equal(subSelect.getSelection(), observation.get(Observation_.observationTime))
    );
    TypedQuery<Observation> typedQuery = entityManager.createQuery(query);
    return typedQuery.getResultList();
}

推荐阅读