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我在教条2上遇到了一些麻烦。我想订购一个查询,但一周中的某一天。Doctrine2 select by orderby(case when)
这是什么意思:
如果这里给出的日期当天是Tuesday
,我想拥有它通过Tuesday, Wednseday, ..., Sunday, Monday
订购。
但是表现可能有多天。
我来到这里的代码确实被
public function getValidPerformancesByDay2($date, $max = 25, $from, $asSql = false){
$myday = intVal(date('w',strtotime($date)));
$q = $this->getEntityManager()
->createQueryBuilder()
->select('DISTINCT textdesc,
CASE WHEN (perfdays.id < '. $myday .') THEN perfdays.id + 8
ELSE perfdays.id END AS HIDDEN sortvalue')
->from ('sys4winsegundaBundle:Performance','textdesc')
->join('textdesc.days', 'perfdays')
->where ('textdesc.enddate >= :date')
->andWhere('textdesc.isvalid = true')
->orderBy('sortvalue','ASC')
->setMaxResults($max)
->setFirstResult($from)
->setParameter('date',$date)
;
$query = $q->getQuery();
if ($asSql){
return $query;
}
return $query->getResult();
}
但该订单的招不幸的是,当我看着已经发送它的查询是:
SELECT DISTINCT p0_.id AS id0, p0_.name AS name1, p0_.duration AS duration2,
p0_.addedby AS addedby3, p0_.startdate AS startdate4, p0_.enddate AS enddate5,
p0_.starthour AS starthour6, p0_.flyer AS flyer7, p0_.price AS price8,
p0_.discount AS discount9, p0_.isvalid AS isvalid10,
p0_.archivedon AS archivedon11, p0_.description AS description12,
p0_.weblink AS weblink13, p0_.techinfo AS techinfo14, p0_.slug AS slug15,
CASE WHEN (d1_.id < 4) THEN d1_.id + 8 ELSE d1_.id END AS sclr16,
p0_.place_id AS place_id17, p0_.gallery_id AS gallery_id18 FROM performances
p0_ INNER JOIN performance_day p2_ ON p0_.id = p2_.performance_id
INNER JOIN days d1_ ON d1_.id = p2_.day_id WHERE p0_.enddate >= ? AND
p0_.isvalid = 1 ORDER BY sclr16 ASC OFFSET 0
Parameters: ['2012-08-23']
Time: 5.13 ms
这意味着如果表演一周发生3次,我发生3次。
有人有想法吗?
编辑 我的英语是非常糟糕的,我lltry以不同的方式解释: 嗯,我得到了在不同天发生文艺演出。
我想要做的是按时间顺序排列它们。 但我发送到数据库的方式是与一个开始日期,结束日期,然后它发生在几天(星期二,星期三...)
我的查询做到了这一点(按最近的一个排序),但作为一些表现发生例如在星期三和星期五,我的查询将返回该表演2次(星期三和星期五),而我应该只检索每个表演的发生,但具有相同的顺序(最近的第一个)
那么我得到了不同日子发生的艺术表演。 –