Java HotSpot Virtual Machine

create or replace function add_working_days(
p_days in number,
p_dt in date default trunc(sysdate)
)
return date
as
v_weeks number;
v_adj number;
begin

v_weeks := trunc(p_days/5);

if to_number(to_char(p_dt,'D')) + mod(p_days,5) >= 7 then
v_adj := 2;
else
v_adj := 0;
end if;

return p_dt + 7*v_weeks + v_adj+mod(p_days,5);
end add_working_days;




This is not working correct if the date is SATURDAY, but seems working M-F and Sunday

select trunc(sysdate)+4,add_working_days(6,trunc(sysdate)+4) from dual;
Output : 19-sep2009 29-Sep-2009

create or replace function add_working_days(
p_days in number,
p_dt in date default trunc(sysdate)
)
return date
as
v_weeks number;
v_adj number;
begin

v_weeks := trunc(p_days/5);


if to_number(to_char(p_dt,'D')) + mod(p_days,5) >= 7 then
v_adj := 2;
else
v_adj := 0;
end if;


return p_dt + 7*v_weeks + v_adj+mod(p_days,5);
end add_working_days;





This is not working correct if the date is SATURDAY, but seems working M-F and Sunday


select trunc(sysdate)+4,add_working_days(6,trunc(sysdate)+4) from dual;
Output : 19-sep2009 29-Sep-2009

Hi, Niren,

[This thread|http://forums.oracle.com/forums/thread.jspa?messageID=3351822&#3351822] should give you some ideas. It does it without using PL/SQL, so it's actually a harder problem.

In PL/SQL, you could loop through the days, starting with the start_date parameter and decrementing a counter that is initialized to the other parameter, stopping when the counter reaches 0.
Or you could write a recursive function.

Like this?

SQL> ed
Wrote file afiedt.buf

create or replace function next_working_day(start_date in date,  add_days in number) return date is
return_date date;
begin
return_date:=start_date+add_days;
if (UPPER(to_char(return_date,'FMDAY'))='SATURDAY') then
return_date:=return_date+2;
elsif (UPPER(to_char(return_date,'FMDAY'))='SUNDAY') then
return_date:=return_date+1;
end if;
return return_date;
end;
SQL> /

Function created.

SQL> select next_working_day('10-AUG-2009',10) FROM DUAL;

NEXT_WORK
---------
20-AUG-09

SQL> select next_working_day('11-SEP-2009',1) from dual;

NEXT_WORK
---------
14-SEP-09

SQL> 

-Arun

Tested

Posted after testing

Hi Frant,

Please explain me with an example. If you have the exact function/Procedure than it can really help.

Thanks
Niren

Hi, Niren,

The first respondent gave a better function for testing for work days (Monday through Friday). If there's a bug with Saturdays, I'm sure that can fixed. Also,that function depends on your NLS settings. If that's an issue for you, that can be fixed, too. The idea that there are 5 work days in any 7 consecutive days will cut down the execution time considerable.

A function like I suggested would be useful if you had to consider holidays as well as weekends.
Here's an untested example:

CREATE OR REPLACE FUNCTION  work_days_away
(      in_start_date	    DATE
,      in_day_cnt	    NUMBER
)
RETURN	DATE
DETERMINISTIC
IS
(
	IF  in_start_date - TRUNC (in_start_date, 'IW') >= 5
	OR  is_holiday (in_start_date)
	THEN
		RETURN  work_days_away ( in_start_date + 1
				       , in_day_cnt
				       );
	ELSIF  in_day_cnt >= 1
	THEN
		RETURN  work_days_away ( in_start_date + 1
				       , in_day_cnt    - 1
				       );
	ELSE
		RETURN  in_start_date;
	END IF;
END	work_days_away;

This use the date format 'IW', which does not depend on NLS settings, rather than 'D', which does.

Hi Arun,

The code example where 10-AUG-2009 + 10 days should return 22 as per the need but your code is not considering the saturday and sunday falling on 15,16 Aug 2009.

Thanks
Niren

Hi,

I was under the impression that, after adding the days, if the result falls on saturday / sunday, it should consider that and return the next working day i.e monday.

So, 10-AUG-2009 + 10 = 20-AUG-2009 which falls on Thursday.
11-SEP-2009 + 1 = 12-SEP-2009 which falls on Saturday. So, the next working day is Monday which is 14-Sep-2009.

Please let me know if my understanding about your requirement is not correct.

-Arun

Hi Arun,

Basically the need is that any number of days added to a date should return date excluding all saturdays and sundays falling in between.
It would be great if you could help me on this .

Thanks
Niren
+919873314561

How about this?

create or replace function add_working_days(
p_days in number,
p_dt in date default trunc(sysdate)
)
return date
as
v_weeks number;
v_adj number;
begin

v_weeks := trunc(p_days/5);

if to_number(to_char(p_dt,'D')) + mod(p_days,5) <= 6 then
v_adj := 1;
elsif to_number(to_char(p_dt,'D')) + mod(p_days,5) = 7 then
v_adj := 2;
else
v_adj := 0;
end if;

return p_dt + 7*v_weeks + v_adj+mod(p_days,5);
end add_working_days;

-Arun

Untested

Dear User,
Simple Select can be used to get the desired result.

Solution is :
select decode(to_char((<your_date> + <days_to_add>),'FMDAY'),'SATURDAY',(((<your_date> + <days_to_add>) +2 ,'SUNDAY' ,(((<your_date> + <days_to_add> ) +1, (((<your_date> + <days_to_add> )) from dual;

Examples :
For sysdate as 16 sep 09

Case 1 : Sunday
SQL> select decode(to_char((sysdate + 4),'FMDAY'),'SATURDAY',(sysdate + 4 ) +2 ,'SUNDAY' ,(sysdate + 4 ) +1, (sysdate + 4 )) from dual;

DECODE(TO
---------
21-SEP-09

Case 2 : Saturday
SQL> select decode(to_char((sysdate + 3),'FMDAY'),'SATURDAY',(sysdate + 3 ) +2 ,'SUNDAY' ,(sysdate + 3 ) +1, (sysdate + 3 )) from dual;

DECODE(TO
---------
21-SEP-09


Case 2 : Week day
SQL> select decode(to_char((sysdate + 2 ),'FMDAY'),'SATURDAY',(sysdate + 2 ) +2 ,'SUNDAY' ,(sysdate + 2 ) +1, (sysdate + 2 )) from dual;

DECODE(TO
---------
18-SEP-09


Revert me back in case of any confusion .

I've written such a logic some years ago. I think you have already given some usable results. Just a few comments from my own experience. There are special scenarios you should think about, how you want to handle them.

1) You add a weekend as date and add 0. Should the next monday be returned or should it stay at the weekend day?
2) Is the number parameter the number of workdays added or the number of days? Like: What should the result be if you add 20 days, will all weekends be skipped? You already gave a clear definition, but it should be specifically mentioned that the number to add is not a number of working days.
3) At a later point you might want to include holidays too (from a holiday table, for example)
4) What about negative numbers? Can your function also calculate back in time? Do you want the friday in this case or the next monday?

Hi Friends,

Thanks for the reply, i have developed the solution.
Sending the code for the benefit of others.

CREATE OR REPLACE FUNCTION NIREN_ADD_DAYS (
p_date DATE,
p_days NUMBER
)
RETURN DATE
IS

v_date DATE;
v_count NUMBER;

BEGIN

SELECT
DECODE(
RTRIM(TO_CHAR(p_date, 'DAY')),
'SATURDAY',
NEXT_DAY(p_date, 'SUNDAY'),
NEXT_DAY((p_date - 7), 'SUNDAY')
)
INTO
v_date
FROM
dual;

v_count := p_date - v_date;

IF (v_count = -1)
THEN

v_count := 0;

END IF;

v_count := v_count + p_days;

IF (v_count > 5)
THEN

v_count := v_count + (FLOOR(v_count / 5.1) * 2);

END IF;

RETURN (v_date + v_count);

END NIREN_ADD_DAYS;
/