Query for the longest duration of consecutive TRUE [duplicate] - sql

Question

I have the following table in SQL Server. I would like to find the longest duration for the machine running.
Row
DateTime
Machine On
1
9/22/2022 8:20
1
2
9/22/2022 9:10
0
3
9/22/2022 10:40
1
4
9/22/2022 10:52
0
5
9/22/2022 12:30
1
6
9/22/2022 14:30
0
7
9/22/2022 15:00
1
8
9/22/2022 15:40
0
9
9/22/2022 16:25
1
10
9/22/2022 16:55
0
In the example above, the longest duration for the machine is ON is 2 hours using rows 5 and 6. What would be the best SQL statement that can provide the longest duration given a time range?
Desired Result:
60 minutes
I have looked into the LAG Function and the LEAD Function in SQL.

Answer 1

Here's another way that uses traditional gaps & islands methodology:
WITH src AS
(
SELECT Island, mint = MIN([Timestamp]), maxt = MAX([Timestamp])
FROM
(
SELECT [Timestamp], Island =
ROW_NUMBER() OVER (ORDER BY [Timestamp]) -
ROW_NUMBER() OVER (PARTITION BY Running ORDER BY [Timestamp])
FROM dbo.Machine_Status
) AS x GROUP BY Island
)
SELECT TOP (1) delta =
(DATEDIFF(second, mint, LEAD(mint,1) OVER (ORDER BY island)))
FROM src ORDER BY delta DESC;
Example db<>fiddle based on the sample data in your new duplicate.

Answer 2

If this is really your data, you can simply use INNER JOIN and DATEDIFF:
SELECT MAX(DATEDIFF(MINUTE, T1.[DateTime], T2.[DateTime]))
FROM [my_table] T1
INNER JOIN [my_table] T2
ON T1.[Row] + 1 = T2.[Row];

Answer 3

This is a gaps and islands problem, one option to solve it is to use a running sum that increased by 1 whenever a machine_on = 0, this will define unique groups for consecutive 1s followed by 0.
select top 1 datediff(minute, min([datetime]), max([datetime])) duration
from
(
select *,
sum(case when machine_on = 0 then 1 else 0 end) over (order by datetime desc) grp
from table_name
) T
group by grp
order by datediff(minute, min([datetime]), max([datetime])) desc
See demo

Answer 4

This is a classic Gaps and Islands with a little twist Adj
Example
Select Top 1
Row1 = min(row)
,Row2 = max(row)+1
,TS1 = min(TimeStamp)
,TS2 = dateadd(SECOND,max(Adj),max(TimeStamp))
,Dur = datediff(Second,min(TimeStamp),max(TimeStamp)) + max(Adj)
From (
Select *
,Grp = row_number() over( partition by Running order by TimeStamp) - row_number() over (order by timeStamp)
,Adj = case when Running=1 and lead(Running,1) over (order by timestamp) = 0 then datediff(second,TimeStamp,lead(TimeStamp,1) over (order by TimeStamp) ) else 0 end
From Machine_Status
) A
Where Running=1
Group By Grp
Order By Dur Desc
Results
Row1 Row2 TS1 TS2 Dur
8 12 2023-01-10 08:25:30.000 2023-01-10 08:28:55.000 205