Lag functions and SUM - sql

Question

I need to get the list of users that have been offline for at least 20 min every day. Here's my data
I have this starting query but am stuck on how to sum the difference in offline_mins i.e. need to add "and sum(offline_mins)>=20" to the where clause
SELECT
userid,
connected,
LAG(recordeddt) OVER(PARTITION BY userid
ORDER BY userid,
recordeddt) AS offline_period,
DATEDIFF(minute, LAG(recordeddt) OVER(PARTITION BY userid
ORDER BY userid,
recordeddt),recordeddt) offline_mins
FROM device_data where connected=0;
My expected results :
Thanks in advance.

Answer 1

This reads like a gaps-and-island problem, where you want to group together adjacent rows having the same userid and status.
As a starter, here is a query that computes the islands:
select userid, connected, min(recordeddt) startdt, max(lead_recordeddt) enddt,
datediff(min(recordeddt), max(lead_recordeddt)) duration
from (
select dd.*,
row_number() over(partition by userid order by recordeddt) rn1,
row_number() over(partition by userid, connected order by recordeddt) rn2,
lead(recordeddt) over(partition by userid order by recordeddt) lead_recordeddt
from device_data dd
) dd
group by userid, connected, rn1 - rn2
Now, say you want users that were offline for at least 20 minutes every day. You can breakdown the islands per day, and use a having clause for filtering:
select userid
from (
select recordedday, userid, connected,
datediff(min(recordeddt), max(lead_recordeddt)) duration
from (
select dd.*, v.*,
row_number() over(partition by v.recordedday, userid order by recordeddt) rn1,
row_number() over(partition by v.recordedday, userid, connected order by recordeddt) rn2,
lead(recordeddt) over(partition by v.recordedday, userid order by recordeddt) lead_recordeddt
from device_data dd
cross apply (values (convert(date, recordeddt))) v(recordedday)
) dd
group by convert(date, recordeddt), userid, connected, rn1 - rn2
) dd
group by userid
having count(distinct case when connected = 0 and duration >= 20 then recordedday end) = count(distinct recordedday)

Answer 2

As noted this is a gaps and island problem. This is my take on it using a simple lag function to create groups, filter out the connected rows and then work on the date ranges.
CREATE TABLE #tmp(ID int, UserID int, dt datetime, connected int)
INSERT INTO #tmp VALUES
(1,1,'11/2/20 10:00:00',1),
(2,1,'11/2/20 10:05:00',0),
(3,1,'11/2/20 10:10:00',0),
(4,1,'11/2/20 10:15:00',0),
(5,1,'11/2/20 10:20:00',0),
(6,2,'11/2/20 10:00:00',1),
(7,2,'11/2/20 10:05:00',1),
(8,2,'11/2/20 10:10:00',0),
(9,2,'11/2/20 10:15:00',0),
(10,2,'11/2/20 10:20:00',0),
(11,2,'11/2/20 10:25:00',0),
(12,2,'11/2/20 10:30:00',0)
SELECT UserID, connected,DATEDIFF(minute,MIN(DT), MAX(DT)) OFFLINE_MINUTES
FROM
(
SELECT *, SUM(CASE WHEN connected <> LG THEN 1 ELSE 0 END) OVER (ORDER BY UserID,dt) grp
FROM
(
select *, LAG(connected,1,connected) OVER(PARTITION BY UserID ORDER BY UserID,dt) LG
from #tmp
) x
) y
WHERE connected <> 1
GROUP BY UserID,grp,connected
HAVING DATEDIFF(minute,MIN(DT), MAX(DT)) >= 20