Avoiding PHP workaround for double GROUP BY sql statement - php

Question

I would like to make a SQL query of counted grouped results. The only way I was able to do it was with the help of PHP. But I am sure that it is possible to do it in one SQL statement.
The table's structure is the following:
id, user_hash, timestamp
And my little PHP workaround looks like this:
$sql = 'SELECT count(id) AS \'total clicks\' FROM stats WHERE timestamp > UNIX_TIMESTAMP(NOW() - INTERVAL '.$num_days.' DAY) GROUP BY user_hash ORDER BY count(id) DESC';
$result = $mysqli->query($sql);
$count = array();
while ($row = mysqli_fetch_assoc($result)) {
if (isset($count[$row['total clicks']])) {
$count[$row['total clicks']]++;
} else {
$count[$row['total clicks']] = 1;
}
}
foreach ($count as $k => $v) {
echo $v." users clicked ".$k." times\n";
}
I am probably going to kick myself how easy it is, but I just can't seem to get to an all-in-one SQL solution. :)

Answer 1

Please try:
$sql = 'SELECT
A.`user_times_clicked`,
COUNT(A.user_hash) AS `user_count`
FROM(
SELECT
count(id) AS `user_times_clicked`,
user_hash
FROM
stats
WHERE
timestamp > UNIX_TIMESTAMP(NOW() - INTERVAL '.$num_days.' DAY)
GROUP BY
user_hash
) AS A
GROUP BY
A.`user_times_clicked`
ORDER BY
COUNT(A.user_hash) DESC'
I hope it helps.