我正在尝试编写一个 CriteriaQuery 它将查询每个城市的最新观察结果。城市由city_code字段定义,最新记录由observation_time字段定义。
我可以轻松地用普通 SQL 编写它,但我无法理解如何使用 jpa criteria api 来实现它。
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`
最佳答案
当您对宽松的效率持开放态度时,这是可能的。 因此,首先让我们将查询转换为逻辑等效查询:
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);
然后让我们将其转换为 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();
}
关于java - 具有聚合内连接的 JPA Criteria 查询,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/58323463/