U kunt MySQL-variabelen gebruiken om de lopende som van momenteel ingelogde bezoekers te berekenen en vervolgens het maximum te krijgen:
SET @logged := 0;
SET @max := 0;
SELECT
idLoginLog, type, time,
(@logged := @logged + IF(type, 1, -1)) as logged_users,
(@max := GREATEST(@max, @logged))
FROM logs
ORDER BY time;
SELECT @max AS max_users_ever;
(SQL Fiddle )
Bewerken: Ik heb ook een suggestie hoe om te gaan met gebruikers die niet expliciet zijn uitgelogd. Stel dat u overweegt dat een gebruiker na 30 minuten automatisch is uitgelogd:
SET @logged := 0;
SET @max := 0;
SELECT
-- Same as before
idLoginLog, type, time,
(@logged := @logged + IF(type, 1, -1)) AS logged_users,
(@max := GREATEST(@max, @logged)) AS max_users
FROM ( -- Select from union of logs and records added for users not explicitely logged-out
SELECT * from logs
UNION
SELECT 0 AS idLoginnLog, l1.username, ADDTIME(l1.time, '0:30:0') AS time, 0 AS type
FROM -- Join condition matches log-out records in l2 matching a log-in record in l1
logs AS l1
LEFT JOIN logs AS l2
ON (l1.username=l2.username AND l2.type=0 AND l2.time BETWEEN l1.time AND ADDTIME(l1.time, '0:30:0'))
WHERE
l1.type=1
AND l2.idLoginLog IS NULL -- This leaves only records which do not have a matching log-out record
) AS extended_logs
ORDER BY time;
SELECT @max AS max_users_ever;
(Fiddle )