Question

以下是短篇小说：

我有3张桌子（1张适合多种关系），这是新闻和俱乐部。

新闻可能与很多俱乐部有关。它也可能没有关系。

例如;我有3个俱乐部的消息。所以，我必须选择我的形式的俱乐部。另一个例子;我有一般公众的新闻。所以，我们不会选择任何俱乐部。

因此;我想用singleScalarResult计算非相关记录。

以下是我获取结果的查询：

// Get entity manager
$em = $this->getDoctrine()->getManager();

// Create query builder
$qb = $em->createQueryBuilder();
$qb
->select('n')
->from('ATLClubBundle:News', 'n')
->where(
    $qb->expr()->isNull('n.deletedAt'),
    $qb->expr()->eq("n.newsType", ":type")
)
->setParameter("type", $newsType);

if (!$slug) {

    // if there is no slug choose both relational and non-relational news

} else if($slug == 'district') {

    // if slug is defined and slug is district get only non-relational news. This is working well with using having but if there any better way to do it, please let me know.
    $qb
        ->leftJoin("n.clubs", "c")
        ->andWhere(
             $qb->expr()->eq('n.showAll', 1)
        )
        ->having('COUNT(c.id) = 0')
        ->groupBy('n.id');
} else {
    // else, just get related news with clubs.
    $qb->innerJoin("n.clubs", "c");
}

    $qb
        ->setFirstResult($offset)
        ->setMaxResults($limit);

此代码块可以很好地获取结果。

当我尝试计算分页结果时;

$counter = $em->createQueryBuilder();
$counter
    ->select('COUNT(n.id)')
    ->from('ATLClubBundle:News', 'n')
    ->where(
        $counter->expr()->isNull('n.deletedAt')
    );

if(!$slug) {
    // same stuff as above
} else if($slug == 'district') {
    $counter
        ->leftJoin("n.clubs", "c", "WITH")
        ->andWhere(
            $counter->expr()->eq('n.showAll', 1)
        );
    // not working, throws exception with having or group by.
} else {
    $counter->innerJoin("n.clubs", "c");
}

$result = $counter->getQuery()->getSingleScalarResult();

我收到错误（因为拥有或分组而引发异常）

Symfony2 Doctrine Count ManyToMany非相关记录

0 个答案: