MS SQL GROUPED SUM - sql

Question

I currently have an MS SQL query which calculates the length of time that each User has been logged in to a system during one day. The table I am extracting this information from records each log in/log out as a separate record. Currently, my MS SQL code is as follows:
SELECT
CAST(DateTime As Date),
UserID,
MIN(DateTime),
MAX(DateTime),
DATEDIFF(SS, MIN(DateTime), MAX(DateTime))
FROM
LoginLogoutData
WHERE
CAST(DateTime AS DATE) = '01/01/2015'
GROUP BY
CAST(DateTime As Date),
UserID
This works as required and creates a table similar to the below.
Date UserID FirstLogIn FinalLogOut LoggedInTime
......... ...... .......... ............ ............
01/01/2015 ABC 07:42:57 14:57:13 26056
01/01/2015 DEF 07:45:49 13:57:56 22326
This works fine for one day's-worth of data. However, if I wanted to calculate the length of time that someone was logged into the system for during a larger date range, e.g. a week or month, this would not work; it would calculate the length of time between the user's log in on the first day and their log out on the final day.
Basically, I would like my code to calculate (Max(DateTime) - Min(DateTime)) FOR EACH DAY then sum all these values together into one simple table grouped only by UserId. I would then be able to set my date range as I please and receive the correct results.
So I would have a table as follows:
UserId LoggedInTime
........ .............
ABC 563287
DEF 485823
GEH 126789
I assume I need to use a GROUP BY within the MIN() function but I don't have much experience with this yet.
Does anyone have any experience with this? Any help would be greatly appreciated.
Thank you.

Answer 1

First you need to aggregate by date, and then by larger units of time. For instance, for the year to date:
SELECT UserId, SUM(diffsecs)
FROM (SELECT CAST(DateTime As Date) as thedate, UserID,
DATEDIFF(second, MIN(DateTime), MAX(DateTime)) as diffsecs
FROM LoginLogoutData
GROUP BY CAST(DateTime As Date), UserID
) ud
WHERE thedate between '2015-01-01' and getdate();

Answer 2

You can use another group by statement to the first query like below:
Select UserID,SUM(LoggedTime)
FROM
(
SELECT CAST(DateTime As Date),
UserID, MIN(DateTime),
MAX(DateTime),
DATEDIFF(SS, MIN(DateTime), MAX(DateTime)) AS LoggedTime
FROM LoginLogoutData
WHERE CAST(DateTime AS DATE) = '01/01/2015'
GROUP BY CAST(DateTime As Date), UserID
) As temp
GROUP BY UserID
Here you can change the where clause to match the data range. It will first calculate logged time for each day and then get the sum of all days per user.

Answer 3

Here is an example of how to do this with working sample data and the T-SQL included.
-- original table you described
CREATE TABLE LoginLogoutData (UserID int, DateTime DateTime)
GO
-- clean any previous sample records
TRUNCATE TABLE LoginLogOutData
/*local variables for iteration*/
DECLARE #i int = 1
DECLARE #n int
DECLARE #entryDate DateTime = GETDATE()
--populate the table with some sample data
/* for each of the five sample users, generate sample login and logout
data for 30 days. Each login and logout are simply an hour apart for demo purposes. */
SET NOCOUNT ON
-- iterate over 5 users (userid)
WHILE (#i <= 5)
BEGIN
--set the initial counter for the date loop
SET #n = 1
--dated entry loop
WHILE (#n <= 30)
BEGIN
-- increment to the next day
SET #entryDate = DateAdd(dd,#n,GETDATE())
--logged in entry
INSERT INTO LoginLogoutData (DateTime, UserID)
SELECT #entryDate,#i
-- logged out entry
INSERT INTO LoginLogoutData (DateTime, UserID)
SELECT DateAdd(hh,1,#entryDate),#i
--increment counter
SET #n = #n+1
END
--increment counter
SET #i=#i+1
END
GO
/* demonstrate that for each user each day has entries and that
the code calculates (Max(DateTime) - Min(DateTime)) FOR EACH DAY
*/
SELECT UserID,
MIN(DateTime) AS LoggedIn,
MAX(DateTime) AS LoggedOut,
DATEDIFF(SS, MIN(DateTime), MAX(DateTime)) AS LoginTime
FROM LoginLogoutData
GROUP BY CAST(DateTime As Date), UserID
/*this is a table variable used to support the "sum all these values together into one
simple table grouped only by UserId*/
DECLARE #SummedUserActivity AS TABLE (UserID int, DailyActivity int)
-- group the subtotals from each day per user
INSERT INTO #SummedUserActivity (UserID, DailyActivity)
SELECT UserID, DATEDIFF(SS, MIN(DateTime), MAX(DateTime))
FROM LoginLogoutData
GROUP BY CAST(DateTime As Date), UserID
-- demonstrate the sum of the subtotals grouped by userid
SELECT UserID, SUM(DailyActivity) AS TotalActivity
FROM #SummedUserActivity
GROUP BY UserID

Answer 4

Since you are already calculating LoggedInTime for each date, the following query would be necessary
SELECT USERID,SUM(LoggedInTime) LoggedInTime
FROM YOURTABLE
GROUP BY USERID
UPDATE
Since you have one record for login and the next record for logout(irrespective of date),we can use the concept of SELF JOIN(a table that joins itself) to get logout datetime for corresponding login time.
DECLARE #FROMDATE DATETIME='2014-01-01 07:42:57.000'
DECLARE #TODATE DATETIME='2015-02-01 07:42:57.000'
;WITH CTE AS
(
-- ROW_NUMBER() is taken as logic for self joining.
SELECT ROW_NUMBER() OVER(ORDER BY USERID,[DATETIME]) RNO,*
FROM #TEMP
WHERE [DATETIME] >= #FROMDATE AND [DATETIME] < #TODATE
)
,CTE2 AS
(
-- Since we are using SELF JOIN,we will get the next row's
-- datetime(ie, Logout time for corresponding login time)
SELECT C1.*,C2.[DATETIME] [DATETIME2],
DATEDIFF(SS, C1.[DateTime], C2.[DateTime]) SECONDSDIFF
FROM CTE C1
LEFT JOIN CTE C2 ON C1.RNO=C2.RNO-1
WHERE C1.RNO%2 <> 0
-- Since we get the next row's datetime in current row,
-- we omit each logout time's row
)
SELECT USERID,SUM(SECONDSDIFF) SECONDSDIFF
FROM CTE2
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