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/

10-15 18:44