我正在使用queryDSL为用户提供一些来自基础的附加数据:
public List<Tuple> getUsersWithData (final SomeParam someParam) {
QUser user = QUser.user;
QRecord record = QRecord.record;
JPQLQuery = query = new JPAQuery(getEntityManager());
NumberPath<Long> cAlias = Expressions.numberPath(Long.class, "cAlias");
return query.from(user)
.leftJoin(record).on(record.someParam.eq(someParam))
.where(user.active.eq(true))
.groupBy(user)
.orderBy(cAlias.asc())
.list(user, record.countDistinct().as(cAlias));
}
尽管它可以按需工作,但它在SQL中生成了两个COUNT():
SELECT
t0.ID
t0.NAME
to.ACTIVE
COUNT(DISTINCT (t1.ID))
FROM USERS t0 LEFT OUTER JOIN t1 ON (t1.SOME_PARAM_ID = ?)
WHERE t0.ACTIVE = true
GROUP BY t0.ID, to.NAME, t0.ACTIVE
ORDER BY COUNT(DISTINCT (t1.ID))
我想知道是否有可能得到这样的东西:
SELECT
t0.ID
t0.NAME
to.ACTIVE
COUNT(DISTINCT (t1.ID)) as cAlias
FROM USERS t0 LEFT OUTER JOIN t1 ON (t1.SOME_PARAM_ID = ?)
WHERE t0.ACTIVE = true
GROUP BY t0.ID, to.NAME, t0.ACTIVE
ORDER BY cAlias
我无法从文档中了解这一点,请给我一些指导。
最佳答案
QVehicle qVehicle = QVehicle.vehicle;
NumberPath<Long> aliasQuantity = Expressions.numberPath(Long.class, "quantity");
final List<QuantityByTypeVO> quantityByTypeVO = new JPAQueryFactory(getEntityManager())
.select(Projections.constructor(QuantityByTypeVO.class, qVehicle.tipo, qVehicle.count().as(aliasQuantity)))
.from(qVehicle)
.groupBy(qVehicle.type)
.orderBy(aliasQuantity.desc())
.fetch();
select
vehicleges0_.type as col_0_0_, count(vehicleges0_.pk) as col_1_0_
from vehicle vehicleges0_
group by vehicleges0_.type
order by col_1_0_ desc;
我做了类似的事情,但在订购之前我先数了数。查找查询并生成选择。
关于java - QueryDSL-按计数顺序作为别名,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/38522167/