Du kan bruge MySQL-variabler til at beregne den løbende sum af aktuelt loggede besøgende og derefter få det maksimale:
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 )
Rediger: Jeg har også et forslag til, hvordan man håndterer brugere, der ikke udtrykkeligt er logget ud. Lad os sige, at du betragter en bruger som automatisk logget ud efter 30 minutter:
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;
(Violin )