Comment calculer les heures écoulées en tenant compte des heures de travail dans Oracle SQL ?

SELECT task,
       start_time,
       end_time,
       ROUND(
         (
           -- Calculate full weeks difference from start of ISO weeks.
           ( TRUNC( end_time, 'IW' ) - TRUNC( start_time, 'IW' ) ) * (10/24) * (6/7)
           -- Add full days for the final week.
           + LEAST( TRUNC( end_time ) - TRUNC( end_time, 'IW' ), 6 ) * (10/24)
           -- Subtract full days from days of the week before the start date.
           - LEAST( TRUNC( start_time ) - TRUNC( start_time, 'IW' ), 6 ) * (10/24)
           -- Add hours of final day
           + LEAST( GREATEST( end_time - TRUNC( end_time ) - 8/24, 0 ), 10/24 )
           -- Subtract hours of the day before the range starts.
           - LEAST( GREATEST( start_time - TRUNC( start_time ) - 8/24, 0 ), 10/24 )
         )
         -- Multiply to give minutes instead of fractions of full days.
         * 24,
         15 -- Number of decimal places
       ) AS work_day_hours_diff
FROM   your_table;

SELECT task,
       COALESCE( SUM( end_time - start_time ), 0 ) * 24 AS total_hours
FROM   (
  SELECT task,
         GREATEST( t.start_time, d.column_value + INTERVAL '8' HOUR ) AS start_time,
         LEAST( t.end_time, d.column_value + INTERVAL '18' HOUR ) AS end_time
  FROM   your_table t
         LEFT OUTER JOIN
         TABLE(
           CAST(
             MULTISET(
               SELECT TRUNC( t.start_time + LEVEL - 1 )
               FROM   DUAL
               WHERE  TRUNC( t.start_time + LEVEL - 1 ) - TRUNC( t.start_time + LEVEL - 1, 'iw' ) < 6
               CONNECT BY TRUNC( t.start_time + LEVEL - 1 ) < t.end_time
             ) AS SYS.ODCIDATELIST
           )
         ) d
         ON (   t.end_time   > d.column_value + INTERVAL  '8' HOUR
            AND t.start_time < d.column_value + INTERVAL '18' HOUR )
)
GROUP BY task;