group subquery counts in mysql statement? - php

Question

Wondering if this is possible or not. In the statement below I am getting counts from different tables. This prints our an array of totals for each user_id.
Is there any way to combine these totals so it returns a single array with the totals? I am using subqueries as joins were taking a hit performance-wise so joining is not an option.
$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 = $stmt->fetchAll(PDO::FETCH_ASSOC);
At the moment I am doing something like this with the return to get the result I want:
foreach($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;

Answer 1

Use SUM aggregate function:
$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
");
This query returns one row with total counts and your required result:
$new = $result[0];