我使用queryDSL从基地获得了一些额外的数据用户:QueryDSL - 为了通过数作为别名
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));
}
尽管它是作为理想的工作,它生成SQL两秒():
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:
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
我不明白这个从documen如果可能的话,请给我一些指导。
感谢您的快速回复! Postgres hadles这个没关系。我知道这对性能并不重要,但它让我感到不安,我想做正确的事情。 –