Как да намерите дата от 48 работни часа, с изключение на събота, неделя и празници в Oracle

Можете да направите всичко това в SQL:

WITH dates ( dt, lvl ) AS (
  SELECT CAST( TRUNC( :your_date ) AS DATE ), 0 FROM DUAL
UNION ALL
  SELECT CAST( dt - INTERVAL '1' DAY AS DATE ),
         CASE
           WHEN ( dt - INTERVAL '1' DAY ) - TRUNC( dt - INTERVAL '1' DAY, 'IW' ) >= 5
           OR   hol_date IS NOT NULL
           THEN lvl
           ELSE lvl + 1
         END
  FROM   dates d
         LEFT OUTER JOIN
         holidays h
         ON ( d.dt - INTERVAL '1' DAY = h.hol_date )
  WHERE  lvl < 2
)
SELECT *
FROM   dates
PIVOT  ( MAX( dt ) FOR lvl IN ( 1 AS DATE24, 2 AS DATE48 ) );

(Забележка:използване на CAST не би трябвало да е необходимо, но без да получавам ORA-01790: expression must have same datatype as corresponding expression )

Или като функция:

CREATE OR REPLACE FUNCTION F_HOL_CHECK_S_NS (
  i_hol_date DATE,
  i_S_NS     NUMBER
) RETURN DATE
AS
  p_date DATE;
BEGIN
  WITH dates ( dt, lvl ) AS (
    SELECT CAST( TRUNC( i_hol_date ) AS DATE ), 0 FROM DUAL
  UNION ALL
    SELECT CAST( dt - INTERVAL '1' DAY AS DATE ),
           CASE
             WHEN ( dt - INTERVAL '1' DAY ) - TRUNC( dt - INTERVAL '1' DAY, 'IW' ) >= 5
             OR   hol_date IS NOT NULL
             THEN lvl
             ELSE lvl + 1
           END
    FROM   dates d
           LEFT OUTER JOIN
           holidays h
           ON ( d.dt - INTERVAL '1' DAY = h.hol_date )
    WHERE  lvl < i_s_ns
  )
  SELECT dt
  INTO   p_date
  FROM   dates
  WHERE  lvl = i_s_ns;

  RETURN p_date;
END;
/