Question

好的我有2个表，相关结构如下：

**youtubevideo**
id - int
allViews - relational


**views**
id - int
youtubevideo_id - int
viewCount - int
firstFetch - date

与youtubevideo有很多观点相关。这是由Symfony实体中的OneToMany提供的。

现在，我有2个日期，让我们从日期和日期调用它们。我需要选择所有youtubevideos并按照属于它们的viewCount总和对它们进行排序。

我只想在fromDate和toDate之间使用具有firstFetch日期的视图。

所以我想它会像

SELECT y FROM YTScraperBundle:YouTubeVideo y 
JOIN y.allViews v GROUP BY y.id 
ORDER BY SUM(v.viewCount) LIMIT 0, 30; <-- or does this need to be an inline select to the specific element?

我并没有真正看到如何将WHERE v.firstFetch放在fromDate和toDate之间。

更新

$query = $this->em->createQuery(
    "SELECT y, IF(v IS NULL, 0, SUM(v.viewCount)) AS HIDDEN sumviewcount
    FROM YTScraperBundle:YouTubeVideo y
    LEFT JOIN y.allViews v WITH v.firstFetch BETWEEN :fromDate AND :toDate
    GROUP BY y ORDER BY sumviewcount DESC
    "
);
$query->setMaxResults(self::PR_PAGE);
$query->setFirstResult($firstResult);
$query->setParameter('fromDate', $fromDate);
$query->setParameter('toDate', $toDate);

获取错误：

Expected known function, got 'IF'

Answer 1

您必须使用WITH，BETWEEN，并将两个DateTime对象绑定到参数化查询中：

$result = $this->getDoctrine()->getManager()->createQuery('
    SELECT y, 
        CASE WHEN (v IS NULL) THEN 0 ELSE SUM(v.viewCount) END AS HIDDEN sumviewcount
    FROM YTScraperBundle:YouTubeVideo y 
    LEFT JOIN y.allViews v WITH v.firstFetch BETWEEN :fromDate AND :toDate
    GROUP BY y ORDER BY sumviewcount DESC
')->setParameter('fromDate', new \DateTime('12-34-5678 00:00:00')) // Replace this with an ISO date
->setParameter('toDate', new \DateTime('12-34-5678 00:00:00')) // Replace here too
->getResult();

我combined @Matteo's answer concerning the SUM into mine for posterity。对HIDDEN想法感激不尽。

Answer 2

对于排序，您可以使用HIDDEN Doctrine2功能（自Doctrine 2.2 +以来可用）。

SELECT y, 
       SUM(v.viewCount) AS HIDDEN sumviewcount 
FROM YTScraperBundle:YouTubeVideo y 
JOIN y.allViews v  
GROUP BY y
ORDER BY sumviewcount 
LIMIT 0, 30;

查看this文章。

希望这个帮助

DQL复杂查询

2 个答案: