php - 学说 2 选择计数 groupBy

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()

但它返回一组组计数。这个应该怎么写才正确？

终于成功了，但是有点丑

$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

尝试将单个 groupBy 分解为 addGroupBy 函数:

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

然而，这实际上返回相同的东西，因为它只是将结果按这 4 个变量分组到唯一的集合中。您应该改为使用 DISTINCT 选择方法并对结果行进行计数。

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

这是我已知的唯一可行的方法，因为 DQL 不支持任何替代策略(例如 SELECT COUNT(*) FROM (SELECT DISTINCT ... ))