教义2选择计数组

Question

我正在尝试检索具有唯一uid的许多行。

$qb->select('COUNT() as cnt')
            ->from($type, 'c')
            ->groupBy('c.organization, c.process_role, c.domain, c.year')
            ->getQuery()->getSingleScalarResult()

但是它返回一组组计数。我应该怎么写这个正确的？

终于可以了，但是有点难看

$count = $this->_em->createQuery( 'SELECT COUNT(c.id) FROM '.$type.' as c WHERE c.id IN ('
            . 'SELECT c1.id FROM ' . $type . ' c1 '
            . 'GROUP BY c1.organization, c1.process_role, c1.domain, c1.year)')
            ->getSingleScalarResult();

Answer 1

尝试将您的单身groupBy分解为addGroupBy功能：

$qb->select('COUNT(c) as cnt')
            ->from($type, 'c')
            ->groupBy('c.organization')
            ->addGroupBy('c.process_role')
            ->addGroupBy('c.domain')
            ->addGroupBy('c.year')
            ->getQuery()->getSingleScalarResult();

但是，这实际上会返回相同的结果，因为它只会将结果按这4个变量分组为唯一的集合。相反，您应该使用DISTINCT选择方法并计算结果行。

$rows = $this->getDoctrine()->getManager()
    ->createQuery(
        'SELECT DISTINCT c.organization, c.process_role, c.domain, c.year FROM ' . $type . ' c'
    )
    ->getArrayResult();
$count = count($rows);

这是我唯一可以使用的已知方法，因为DQL不支持任何替代策略（例如SELECT COUNT(*) FROM (SELECT DISTINCT ... )）