想知道这是否可能。在下面的声明中,我从不同的表中获取计数。这会打印每个 user_id 的总计数组。
有什么方法可以组合这些总计,以便返回包含总计的单个数组吗?我使用子查询是因为连接在性能方面受到了影响,因此连接不是一个选项。
$stmt = $db->prepare("
SELECT
(SELECT COUNT(*) FROM log1 WHERE log1.user_id = users.user_id AS l1,
(SELECT COUNT(*) FROM log2 WHERE log2.user_id = users.user_id AS l2,
(SELECT COUNT(*) FROM log3 WHERE log3.user_id = users.user_id AS l3,
(SELECT COUNT(*) FROM log4 WHERE log4.user_id = users.user_id AS l4
FROM computers
INNER JOIN users
ON users.computer_id = computers.computer_id
WHERE computers.account_id = :cw_account_id AND computers.status = :cw_status
");
$binding = array(
'cw_account_id' => $_SESSION['user']['account_id'],
'cw_status' => 1
);
$stmt->execute($binding);
$result = $stmt->fetchAll(PDO::FETCH_ASSOC);
目前我正在做这样的事情,并返回以获得我想要的结果:
foreach($result as $key)
{
$new['l1'] = $new['l1'] + $key['l1'];
$new['l2'] = $new['l2'] + $key['l2'];
$new['l3'] = $new['l3'] + $key['l3'];
$new['l4'] = $new['l4'] + $key['l4'];
}
return $new;
最佳答案
$stmt = $db->prepare("
SELECT
SUM((SELECT COUNT(*) FROM log1 WHERE log1.user_id = users.user_id)) AS l1,
SUM((SELECT COUNT(*) FROM log2 WHERE log2.user_id = users.user_id)) AS l2,
SUM((SELECT COUNT(*) FROM log3 WHERE log3.user_id = users.user_id)) AS l3,
SUM((SELECT COUNT(*) FROM log4 WHERE log4.user_id = users.user_id)) AS l4
FROM computers
INNER JOIN users
ON users.computer_id = computers.computer_id
WHERE computers.account_id = :cw_account_id AND computers.status = :cw_status
");
此查询返回一行,其中包含总计数和您所需的结果:
$new = $result[0];
关于php - mysql语句中的组子查询计数?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/26229784/