I'm looking for the Oracle (10g) equivalent of:
DATEADD(weekday, -3, GETDATE()) from T-SQL (SQL Server) . This subtracts 3 weekdays from the current date. I'm not concerned about holidays or anything like that (and I can truncate the time part off myself). Just excluding weekends is fine.
It looks like you need to create a UDF.
CREATE OR REPLACE FUNCTION business_date (start_date DATE, days2add NUMBER) RETURN DATE IS Counter NATURAL := 0; CurDate DATE := start_date; DayNum POSITIVE; SkipCntr NATURAL := 0; Direction INTEGER := 1; -- days after start_date BusinessDays NUMBER := Days2Add; BEGIN IF Days2Add < 0 THEN Direction := - 1; -- days before start_date BusinessDays := (-1) * BusinessDays; END IF; WHILE Counter < BusinessDays LOOP CurDate := CurDate + Direction; DayNum := TO_CHAR( CurDate, 'D'); IF DayNum BETWEEN 2 AND 6 THEN Counter := Counter + 1; ELSE SkipCntr := SkipCntr + 1; END IF; END LOOP; RETURN start_date + (Direction * (Counter + SkipCntr)); END business_date; Courtesy of Larry Benton, from here.
It can be done without a PL/SQL function. Just subtract a different number of days depending on the day of the week:
select trunc(sysdate) - case to_char(sysdate, 'D') when '4' then 3 -- thursday minus 3 days when '5' then 3 -- friday minus 3 days when '6' then 4 -- saturday minus 4 days else 5 -- all other days minus 5 days end from dual; When you have to do this for e.g. 12 days back, it would look like:
select trunc(sysdate) - case to_char(sysdate, 'D') when '1' then 18 -- mondays minus 18 days (incl. 3 weekends) when '2' then 18 -- tuesdays minus 18 days (incl. 3 weekends) when '6' then 17 -- saturdays minus 17 days (incl. 2 weekends and a saturday) else 16 -- all others minus 16 days (incl. 2 weekends) end from dual; Please note that day of week depends on the NLS_TERRITORY of your database (in America day 1 is Sunday, in most others day 1 is Monday).