Calculate only working hours between dates in ORACLE - oracle

Question

Hello everyone I have written an oracle query which is calculating working according to 8 hours, but I want according to 8.5 hours result, there is a minor change but I am not getting it please help. Now according to the start and end date, it should return 8.5 working house, but it is returning 8 working hours please assist.
Query
with dates as (
select to_date('20-oct-2022 09:00:00','dd-mon-yyyy hh24:mi:ss') start_dt,
to_date('20-oct-2022 17:30:00','dd-mon-yyyy hh24:mi:ss') end_dt
from dual
),
-- get work hours for each date
t as (
select case level
when 1 then greatest(start_dt,trunc(start_dt) + 8 / 24)
else trunc(start_dt) + level - 16 / 24
end start_dt,
case connect_by_isleaf
when 1 then least(end_dt,trunc(end_dt) + 17 / 24)
else trunc(start_dt) + level - 7 / 24
end end_dt
from dates
connect by level <= trunc(end_dt) - trunc(start_dt) + 1
)
select sum(greatest(end_dt - start_dt,0)) * 24 work_hours
from t
where trunc(start_dt) - trunc(start_dt,'iw') < 5

Answer 1

You do not need to generate all the days; you can directly calculate the number of hours:
SELECT start_dt,
end_dt,
ROUND(
(
-- Calculate the full weeks difference from the start of ISO weeks.
( TRUNC( end_dt, 'IW' ) - TRUNC( start_dt, 'IW' ) ) * 8.5 * (5/7)
-- Add the full days for the final week.
+ LEAST( TRUNC( end_dt ) - TRUNC( end_dt, 'IW' ), 5 ) * 8.5
-- Subtract the full days from the days of the week before the start date.
- LEAST( TRUNC( start_dt ) - TRUNC( start_dt, 'IW' ), 5 ) * 8.5
-- Add the hours of the final day
+ CASE
WHEN end_dt - TRUNC( end_dt, 'IW' ) < 5 -- Weekday
THEN LEAST(
GREATEST(
end_dt - (TRUNC( end_dt ) + INTERVAL '09:00' HOUR TO MINUTE),
0
) * 24,
8.5
)
ELSE 0
END
-- Subtract the hours of the day before the range starts.
- CASE
WHEN start_dt - TRUNC( start_dt, 'IW' ) < 5 -- Weekday
THEN LEAST(
GREATEST(
start_dt - (TRUNC( start_dt ) + INTERVAL '09:00' HOUR TO MINUTE),
0
) * 24,
8.5
)
ELSE 0
END
),
15 -- Number of decimal places
) AS work_hours_diff
FROM dates;
Which, for the sample data:
CREATE TABLE dates (start_dt, end_dt) AS
SELECT DATE '2022-10-20' + INTERVAL '09:00:00' HOUR TO SECOND,
DATE '2022-10-20' + INTERVAL '17:30:00' HOUR TO SECOND
FROM DUAL
UNION ALL
SELECT DATE '2022-10-20' + INTERVAL '10:00:00' HOUR TO SECOND,
DATE '2022-10-21' + INTERVAL '17:30:00' HOUR TO SECOND
FROM DUAL
UNION ALL
SELECT DATE '2022-11-19' + INTERVAL '23:46:00' HOUR TO SECOND,
DATE '2022-11-21' + INTERVAL '12:06:00' HOUR TO SECOND
FROM DUAL
UNION ALL
SELECT DATE '2022-11-18' + INTERVAL '17:30:00' HOUR TO SECOND,
DATE '2022-11-21' + INTERVAL '09:00:00' HOUR TO SECOND
FROM DUAL
UNION ALL
SELECT DATE '2022-11-19' + INTERVAL '12:26:45' HOUR TO SECOND,
DATE '2022-11-21' + INTERVAL '11:02:15' HOUR TO SECOND
FROM DUAL
UNION ALL
SELECT DATE '2022-11-21' + INTERVAL '11:02:15' HOUR TO SECOND,
DATE '2022-11-19' + INTERVAL '12:26:45' HOUR TO SECOND
FROM DUAL;
Outputs:
START_DT
END_DT
WORK_HOURS_DIFF
2022-10-20 09:00:00 (THU)
2022-10-20 17:30:00 (THU)
8.5
2022-10-20 10:00:00 (THU)
2022-10-21 17:30:00 (FRI)
16
2022-11-19 23:46:00 (SAT)
2022-11-21 12:06:00 (MON)
3.1
2022-11-18 17:30:00 (FRI)
2022-11-21 09:00:00 (MON)
0
2022-11-19 12:26:45 (SAT)
2022-11-21 11:02:15 (MON)
2.0375
2022-11-21 11:02:15 (MON)
2022-11-19 12:26:45 (SAT)
-2.0375
Note: the negative value is valid as the start date is after the end date for that row.
fiddle