Providing date back based on number interval

Question

Below is a SQL example of what I'm trying to accomplish. The parameter I will be using is what days are available to work on in the format of (0,1,1,1,1,1,0).
The first number represents Sunday and the last is Saturday. If the day has a 0 you have to skip that date, but if the date has a 1 you can use that date. Therefore if the data contained a 0 for Monday if would have to skip Monday when adding days. I can figure out how to put the SQL together to skip Saturday and Sundays but can't figure out how to use the (0,1,1,1,1,1,0) format to determine if work is allowed.
Here is the SQL function I'm using as concept which I need help on how to incorporate it.
~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*
CREATE OR REPLACE Function custom_add_days
(start_date_in date, days_in number)
return date
IS
v_counter number;
v_new_date date;
v_day_number number;
BEGIN
/* This routine will add a specified number of days (ie: days_in) to a date (ie: start_date). */
/* It will skip all weekend days - Saturdays and Sundays */
v_counter := 1;
v_new_date := start_date_in;
/* Loop to determine how many days to add */
while v_counter <= days_in
loop
/* Add a day */
v_new_date := v_new_date + 1;
v_day_number := to_char(v_new_date, 'd');
/* Increment counter if day falls between Monday to Friday */
if v_day_number >= 2 and v_day_number <= 6 then
v_counter := v_counter + 1;
end if;
end loop;
RETURN v_new_date;

Answer 1

Sorry didn't see your last post again. Changed your query a bit once again.
SQL> CREATE OR REPLACE FUNCTION custom_add_days (
  2     start_date_in   DATE,
  3     work_days       VARCHAR2,
  4     days_in         NUMBER
  5  )
  6     RETURN DATE
  7  IS
  8     v_counter          NUMBER;
  9     v_new_date         DATE;
10     v_day_number       NUMBER;
11     v_loop_counter     NUMBER := 0;
12     work_days_status   NUMBER;
13  BEGIN
14  /* This routine will add a specified number of days (ie: days_in) to a date (ie: start_date). *
15  /* It will skip all weekend days - Saturdays and Sundays */
16     v_counter := 1;
17     v_new_date := start_date_in;
18     v_loop_counter := to_number(to_char(start_date_in, 'D'));
19   
20  /* Loop to determine how many days to add */
21     WHILE v_counter <= days_in
22     LOOP
23  /* Add a day */
24        v_new_date := v_new_date + 1;
25        v_day_number := TO_NUMBER(TO_CHAR (v_new_date, 'd'));
26        v_loop_counter := v_loop_counter + 1;
27        work_days_status :=
28           TO_NUMBER (SUBSTR (REPLACE (work_days, ','),
29                              CASE
30                                 WHEN MOD (v_loop_counter, 7) = 0
31                                    THEN 7
32                                 ELSE MOD (v_loop_counter, 7)
33                              END,
34                              1
35                             )
36                     );
37  /* Increment counter if day falls between Monday to Friday */
38        IF  work_days_status = 1
39        THEN
40           v_counter := v_counter + 1;
41        END IF;
42       
43     END LOOP;
44 
45     RETURN v_new_date;
46  END;
47  /
Function created.
SQL> SELECT CUSTOM_ADD_DAYS(sysdate, '1,0,1,1,1,0,1', 3) FROM Dual
  2   /
CUSTOM_AD
05-JUL-09
1 row selected.
SQL>  SELECT CUSTOM_ADD_DAYS(sysdate, '1,0,1,1,1,0,0', 2) FROM Dual
  2   /
CUSTOM_AD
05-JUL-09
1 row selected.
SQL> Hope this helps.
Regards,
Jo
Edited by: Joice John on Jul 1, 2009 7:44 AM