find number of hours gap when no one is working - sql

Question

i have a table view of table
taking scenario of a hotel that works 24/7 non stop, i want to calculate total number of hours in a month when no one is present .
in image you can see we have shift hours in 24hh format, shifts are 8 hours long and different employees may have different week offs , we have date range for which this shift would be valid , i.e. a typical month and we have planned leaves when employee is completely off.
can you suggest oracle SQL to find total number of hours when no employee is working in a month.

Answer 1

You can generate the shifts and then find the shifts where no-one worked:
WITH shifts (shift_start) AS (
SELECT DATE '2022-07-01' + INTERVAL '6' HOUR + INTERVAL '8' HOUR * (LEVEL - 1)
FROM DUAL
CONNECT BY
DATE '2022-07-01' + INTERVAL '6' HOUR + INTERVAL '8' HOUR * (LEVEL - 1)
< DATE '2022-08-01'
)
SELECT s.shift_start
FROM shifts s
WHERE NOT EXISTS(
SELECT 1
FROM work_master w
WHERE w.shift_date_frm <= s.shift_start
AND s.shift_start < w.shift_date_to + INTERVAL '1' DAY
AND ':' || w.week_off_day || ':' NOT LIKE '%:' || TO_CHAR(s.shift_start, 'Dy') || ':%'
AND EXTRACT(HOUR FROM CAST(s.shift_start AS TIMESTAMP)) = w.shift_start
AND (
(
w.vac_date_frm IS NULL
AND w.vac_date_to IS NULL
)
OR NOT (
w.vac_date_frm <= s.shift_start
AND s.shift_start < w.vac_date_to + INTERVAL '1' DAY
)
)
)
Which, for the sample data:
CREATE TABLE work_master (
employee_name,
shift_start,
shift_end,
shift_date_frm,
shift_date_to,
vac_date_frm,
vac_date_to,
week_off_day
) AS
SELECT 'emp1', 22, 6, DATE '2022-07-01', DATE '2022-07-31', NULL, NULL, 'Sat:Sun' FROM DUAL UNION ALL
SELECT 'emp2', 14, 22, DATE '2022-07-01', DATE '2022-07-31', NULL, NULL, 'Sat:Sun' FROM DUAL UNION ALL
SELECT 'emp3', 6, 14, DATE '2022-07-01', DATE '2022-07-31', DATE '2022-07-27', DATE '2022-07-27', 'Sat:Sun' FROM DUAL UNION ALL
SELECT 'emp4', 14, 22, DATE '2022-07-01', DATE '2022-07-31', NULL, NULL, 'Fri:Sat' FROM DUAL UNION ALL
SELECT 'emp5', 22, 6, DATE '2022-07-01', DATE '2022-07-31', NULL, NULL, 'Wed:Thu' FROM DUAL;
Outputs:
SHIFT_START
2022-07-02 06:00:00 (Sat)
2022-07-02 14:00:00 (Sat)
2022-07-03 06:00:00 (Sun)
2022-07-09 06:00:00 (Sat)
2022-07-09 14:00:00 (Sat)
2022-07-10 06:00:00 (Sun)
2022-07-16 06:00:00 (Sat)
2022-07-16 14:00:00 (Sat)
2022-07-17 06:00:00 (Sun)
2022-07-23 06:00:00 (Sat)
2022-07-23 14:00:00 (Sat)
2022-07-24 06:00:00 (Sun)
2022-07-27 06:00:00 (Wed)
2022-07-30 06:00:00 (Sat)
2022-07-30 14:00:00 (Sat)
2022-07-31 06:00:00 (Sun)
If you just want the total hours then:
WITH shifts (shift_start) AS (
SELECT DATE '2022-07-01' + INTERVAL '6' HOUR + INTERVAL '8' HOUR * (LEVEL - 1)
FROM DUAL
CONNECT BY
DATE '2022-07-01' + INTERVAL '6' HOUR + INTERVAL '8' HOUR * (LEVEL - 1)
< DATE '2022-08-01'
)
SELECT COUNT(*) * 8 AS hours_not_worked
FROM shifts s
WHERE NOT EXISTS(
SELECT 1
FROM work_master w
WHERE w.shift_date_frm <= s.shift_start
AND s.shift_start < w.shift_date_to + INTERVAL '1' DAY
AND ':' || w.week_off_day || ':' NOT LIKE '%:' || TO_CHAR(s.shift_start, 'Dy') || ':%'
AND EXTRACT(HOUR FROM CAST(s.shift_start AS TIMESTAMP)) = w.shift_start
AND (
(
w.vac_date_frm IS NULL
AND w.vac_date_to IS NULL
)
OR NOT (
w.vac_date_frm <= s.shift_start
AND s.shift_start < w.vac_date_to + INTERVAL '1' DAY
)
)
)
Which outputs:
HOURS_NOT_WORKED
128
db<>fiddle here