I've been asked to look at a database that records user login and logout activity - 开发者_JS百科there's a column for login time and then another column to record logout, both in OLE format. I need to pull together some information about user concurrency - i.e. how many users were logged in at the same time each day.
Do anyone know how to do this in SQL? I don't really need to know the detail, just the count per day.
Thanks in advance.
Easiest way is to make a times_table from an auxiliary numbers table (by adding from 0 to 24 * 60 minutes to the base time) to get every time in a certain 24-hour period:
SELECT MAX(simul) FROM (
SELECT test_time
,COUNT(*) AS simul
FROM your_login_table
INNER JOIN times_table -- a table/view/subquery of all times during the day
ON your_login_table.login_time <= times_table.test_time AND times_table.test_time <= your_login_table.logout_time
GROUP BY test_time
) AS simul_users (test_time, simul)
I think this will work.
Select C.Day, Max(C.Concurrency) as MostConcurrentUsersByDay
FROM
(
SELECT convert(varchar(10),L1.StartTime,101) as day, count(*) as Concurrency
FROM login_table L1
INNER JOIN login_table L2
ON (L2.StartTime>=L1.StartTime AND L2.StartTime<=L1.EndTime) OR
(L2.EndTime>=L1.StartTime AND L2.EndTime<=L1.EndTime)
WHERE (L1.EndTime is not null) and L2.EndTime Is not null) AND (L1.ID<>L2.ID)
GROUP BY convert(varchar(10),L1.StartTime,101)
) as C
Group BY C.Day
Unchecked... but lose date values, count time between, use "end of day" for still logged in.
This assumes "logintime" is a date and a time. If not, the derived table can be removed (Still need ISNULL though). of course, SQL Server 2008 has "time" to make this easier too.
SELECT
COUNT(*)
FROM
(
SELECT
DATEADD(day, DATEDIFF(day, logintime, 0), logintime) AS inTimeOnly,
ISNULL(DATEADD(day, DATEDIFF(day, logouttime, 0), logintime), '1900-01-01 23:59:59.997') AS outTimeOnly
FROM
mytable
) foo
WHERE
inTimeOnly >= @TheTimeOnly AND outTimeOnly <= @TheTimeOnly
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