Oracle-fråga för att utesluta helger och 18:00 till 21:00

Du kan direkt räkna ut skillnaden i dagar (anpassat från mitt svar här ):

SELECT start_date,
       end_date,
       ROUND(
         (
           -- Calculate the full weeks difference from the start of ISO weeks.
           ( TRUNC( end_date, 'IW' ) - TRUNC( start_date, 'IW' ) ) * (9/24) * (5/7)
           -- Add the full days for the final week.
           + LEAST( TRUNC( end_date ) - TRUNC( end_date, 'IW' ), 5 ) * (9/24)
           -- Subtract the full days from the days of the week before the start date.
           - LEAST( TRUNC( start_date ) - TRUNC( start_date, 'IW' ), 5 ) * (9/24)
           -- Add the hours of the final day
           + LEAST( GREATEST( end_date - TRUNC( end_date ) - 9/24, 0 ), 9/24 )
           -- Subtract the hours of the day before the range starts.
           - LEAST( GREATEST( start_date - TRUNC( start_date ) - 9/24, 0 ), 9/24 )
         )
         -- Multiply to give minutes rather than fractions of full days.
         * 24 * 60
       ) AS work_day_mins_diff
FROM   table_name;

Som för exempeldata:

CREATE TABLE table_name ( start_date, end_date ) AS
SELECT DATE '2020-12-30' + INTERVAL '00' HOUR, DATE '2020-12-30' + INTERVAL '12' HOUR FROM DUAL UNION ALL
SELECT DATE '2020-12-30' + INTERVAL '18' HOUR, DATE '2020-12-30' + INTERVAL '20' HOUR FROM DUAL UNION ALL
SELECT DATE '2020-12-30' + INTERVAL '17:30' HOUR TO MINUTE, DATE '2020-12-30' + INTERVAL '21:30' HOUR TO MINUTE FROM DUAL UNION ALL
SELECT DATE '2021-01-01' + INTERVAL '00' HOUR, DATE '2021-01-04' + INTERVAL '00' HOUR FROM DUAL UNION ALL
SELECT DATE '2021-01-02' + INTERVAL '00' HOUR, DATE '2021-01-04' + INTERVAL '00' HOUR FROM DUAL UNION ALL
SELECT DATE '2020-12-28' + INTERVAL '00' HOUR, DATE '2021-01-04' + INTERVAL '00' HOUR FROM DUAL UNION ALL
SELECT DATE '2020-12-28' + INTERVAL '00' HOUR, DATE '2020-12-29' + INTERVAL '00' HOUR FROM DUAL;

Utgångar:

(Använder ALTER SESSION SET NLS_DATE_FORMAT = 'YYYY-MM-DD HH24:MI:SS (DY)'; )

db<>fiol här