last working day of each week

Hello world
Could someone suggest how to configure Oracle 11g Scheduler so that it would start the last working day of each week? The last business day of the week is not always Friday; exceptional days are configured in separate planners, who are included or excluded in clause repeat_interval, (for example).

The last business day of each month, works well, like this:

BEGIN
DBMS_SCHEDULER. () CREATE_SCHEDULE
schedule_name = > "LAST_SETT_DAY_OF_MONTH"
repeat_interval = > ' FREQ = MONTHLY; BYHOUR = 18; BYMINUTE = 30; BYDAY IS MON, MAR, SEA, GAME, FRI.; BYSETPOS =-1; EXCLUDE = HOLIDAY; INCLUDE = EXCEPTIONAL_WORKING_DAYS; ") ;
END;

But I want to run a similar logic for weeks.

Thanks in advance

OK, now I understand your request.
I think it might be possible for PL/SQL function.
Create the table suite where you put your nonworking days and the additional working days. To create the PL/SQL function and use

CREATE TABLE day_list(
DAY DATE NOT NULL,
TYPE NUMBER(1) NOT NULL,
CONSTRAINT pk_day_list PRIMARY KEY(DAY),
CONSTRAINT ch_day_list_type CHECK(TYPE IN (1,2)));
-- type 1-nonworking day,2-extra working day

CREATE OR REPLACE FUNCTION last_working_day(p_date IN DATE) RETURN DATE AS
  v_retval DATE := TRUNC(p_date);
  v_aux    DATE;
BEGIN
  IF v_retval IS NOT NULL
  THEN
    -- get next friday
    dbms_scheduler.evaluate_calendar_string(calendar_string => 'FREQ=DAILY; BYDAY=FRI',start_date => v_retval,return_date_after => v_retval,next_run_date => v_retval);
    -- check if SUNDAY or SATURDAY is extra working day
    SELECT MAX(DAY)
    INTO   v_aux
    FROM day_list
    WHERE TYPE=2
    AND DAY BETWEEN v_retval+1 AND v_retval+2;
    IF v_aux IS NOT NULL
    THEN
      v_retval := v_aux;
    ELSE
      -- find last working day from monday to friday
      WITH v_1 AS (SELECT v_retval - LEVEL +1 AS DAY
                     FROM dual
                     CONNECT BY LEVEL <=5)
      SELECT MAX(DAY)
      INTO   v_aux
      FROM (SELECT v_1.day, CASE WHEN t.type=1 THEN 0 ELSE 1 END is_working_day
            FROM v_1
            LEFT JOIN day_list t ON v_1.day=t.day AND t.type=1
            )
      WHERE is_working_day=1;
      -- there is a working day
      IF v_aux IS NOT NULL
      THEN
        v_retval := v_aux;
      ELSE
        -- go to next week
        v_retval := last_working_day(v_retval);
      END IF;
    END IF;
  END IF;
  RETURN v_retval;
END;
/

Tags: Database

Similar Questions

I'm looking for the 1st working day of the week.
Hello

I'm looking for the 1st working day of the week except Saturday, Sunday and holidays. I have vacation, the name of the table that contains the columns HolidayDate and HolidayDescription.

Ex:

01/10/2011 is monday (i.e) 1st company day.check if it is in the host table, then return the result as 01/11/2011 other returns the output as a 11-1-2011(which is the first business day of this week).

Can someone tell me the query to solve?
Hello

If dt is any DATE, it is the first day of work in the same week as dt:
```
WITH     all_week     AS
(
     SELECT     TRUNC (dt, 'IW') + LEVEL - 1     AS a_dt
     FROM     dual
     CONNECT BY     LEVEL <= 5
)
SELECT     MIN (a_dt)     AS first_work_dt
FROM     all_week
WHERE     a_dt     NOT IN (
                  SELECT  holiday_dt
                  FROM    holiday
                 )
;
```
This assumes that you have never 5 consecutive days. (Where I work, holidays are always at least a week apart, so that's fine.)
If you have a week's holiday, increase the number of 'magical' in the CONNECT BY clause.

It also means that holiday.holiday_dt = TRUNC (holiday.holiday.dt).

How to view the 3rd working day of each month.

Hello

could someone help me with a proc to display the 3 day working rd of each month?
By day, I mean working Mon - Fri.
Thank you.

Hello

Sorry, I initially responded to the question "what is the * first * work of the month ', which is a little easier.

Here's a way you can code in SQL:

ALTER SESSION     SET NLS_DATE_FORMAT     = 'DD-Mon-YYYY Dy';

WITH     got_dates     AS
(
     SELECT     TRUNC ( ADD_MONTHS (SYSDATE, LEVEL)
                , 'MONTH'
                )     AS a_date
     FROM     dual
     CONNECT BY     LEVEL     <= 12
)
-- The part above just generates some sample dates
-- The part below is the actual query:
--
SELECT    a_date
,       CASE  TO_CHAR ( a_date + 2
                        , 'fmDay'
                  , 'NLS_DATE_LANGUAGE=ENGLISH'     -- If needed
                  )
           WHEN  'Monday'    THEN  4
           WHEN  'Tuesday'     THEN  3
           WHEN  'Saturday'  THEN  4
           WHEN  'Sunday'     THEN  4
                              ELSE  2
       END + a_date     AS third_workday
FROM       got_dates
ORDER BY  a_date
;

Output:

A_DATE            THIRD_WORKDAY
---------------   ---------------
01-Nov-2012 Thu   05-Nov-2012 Mon
01-Dec-2012 Sat   05-Dec-2012 Wed
01-Jan-2013 Tue   03-Jan-2013 Thu
01-Feb-2013 Fri   05-Feb-2013 Tue
01-Mar-2013 Fri   05-Mar-2013 Tue
01-Apr-2013 Mon   03-Apr-2013 Wed
01-May-2013 Wed   03-May-2013 Fri
01-Jun-2013 Sat   05-Jun-2013 Wed
01-Jul-2013 Mon   03-Jul-2013 Wed
01-Aug-2013 Thu   05-Aug-2013 Mon
01-Sep-2013 Sun   04-Sep-2013 Wed
01-Oct-2013 Tue   03-Oct-2013 Thu

Published by: Frank Kulash, October 4, 2012 17:56

Work day of the week of the date of
Hi guys, is a bit common code that allows you to work on the day of the week for a given date? IM pretty new to this and I don't know how to write it and thought since it's probably quite commonly need I may be able to get help here. Thank you.
Day of the week in letters
Select to_char (sysdate, 'day') of double;

Day of week in numbers
Select double to_char(sysdate,'D');

Day of the month in figures
Select to_char (sysdate, 'DD') of double;

PL/SQL function to calculate the non-working days

Hello
I have the following pl/sql function which generates a number of days between two selected dates (i.e. excluding weekends). However, I also need to exclude specific holidays - day of Christmas etc. These holiday dates are stored in a table in our database (11 GR 2) called "HOLIDAY".
How can I integrate the holidays which is held in the table of holidays in the following query to exclude these dates as well? I know how to write a separate funtion pl/sql for a number of days between two dates using SELECT... BUT I can't work out how to bring together them in a single query.
Could someone show me how with the pl/sql / dates below please?
Example of HOLIDAY table below.
Thank you!
TP
```
create or replace
function WORKING_DAYS (pi_start_date in date, pi_end_date in date) return integer

is

v_start_date date :=pi_start_date;
v_end_date date:=pi_end_date;
v_count integer:=0;

begin

while v_start_date <= v_end_date
      loop
                        if to_char(v_start_date,'D') not in ('6','7')                    
                        then
                        v_count := v_count+1;                       
                        end if;
                       
                v_start_date:=v_start_date+1;
               
      end loop;
     
return v_count;

end;
```
```
(select '10-Apr-2013' as NWD from dual union all
select '06-May-2013' from dual union all
select '27-May-2013' from dual union all
select '26-Aug-2013' from dual union all
select '26-Dec-2013' from dual union all
select '25-Dec-2013' from dual union all
select '01-Jan-2014' from dual union all
select '18-Apr-2014' from dual union all
select '21-Apr-2014' from dual union all
select '05-May-2014' from dual union all
select '26-May-2014' from dual union all
select '25-Aug-2014' from dual union all
select '25-Dec-2014' from dual union all
select '26-Dec-2014' from dual) HOLIDAYS
```
Hello

the link is on MOSC, not OTN... Here's a copy:

This function calculates the number of days between two dates, ignoring weekends and holidays (if requested and if the holidays are stored in a table)

I give an example of table 'public_holiday' with sample data, but users must ensure that their table contains the relevant data (all holidays within the maximum range of use of the service)

CREATE TABLE public_holiday (calendar_day, DATE, text VARCHAR2 (30));

FUNCTION to CREATE or REPLACE nb_days (p_date_from IN DATE

p_date_to DATE by DEFAULT TRUNC (sysdate)

, p_public_holidays in CHAR DEFAULT 'Y '.

) RETURN NUMBER

DEFINE AUTHID

AS

/*********************************************************************/

/ * Author: Bruno Vroman * /.

/ * Created: 23-AUG-2012 * /.

/ * Last updated: 23-AUG-2012 * /.

/ * Object: to calculate the number of days between 2 dates, to the exclusion of * /.

/ * Saturday and Sunday, but also "holidays" If the * /.

/ * argument 'p_public_holidays' = 'Y ' * /.

/ * Support: p_date_from<= p_date_to                             ="">

/ * component ' hour min dry "ignored (just counting days) * /.

/* First step:                                                       */

/ * the calendar days between 2 days * /.

/ Remove 2 days for each "week" and 0 or 1 extra day * /.

/ * function to a condition "complex" mix day of first week * /.

/ * and the number of days when full remaining weeks are removed * /.

/ * (set up once for all, for example if there is 3 days and the * /)

(/ * first day is a Thursday, there is 1 "Sat/Sun" to subtract) * /.

/ * Second step: If 'p_public_holidays' = 'Y': other * /.

/ * days, do not count holidays.                         */

/ Holiday everyday are defined in a table "public_holiday" * /.

/ * Note: there may be holidays defined on Saturday/Sunday.       */

/*********************************************************************/

l_result NUMBER;

l_from DATE;

l_to DATE;

l_case tank (4);

BEGIN

l_from: = TRUNC (p_date_from);

l_to: = TRUNC (p_date_to);

l_case: = TO_CHAR (l_from, 'Dy', 'NLS_DATE_LANGUAGE = English').

To_char (MOD (l_to - l_from + 1, 7));

l_result: = l_to - l_from + 1

-TRUNC ((l_to-l_from + 1) / 7) * 2

-CASE

WHEN l_case IN ('Mon6', 'Tue5', 'Wed4', 'Thu3', 'Fri2'

, "Sat1", "Sun1", "Sun2', 'Sun3", "Sun4".

, "Sun5', 'Sun6.

)

THEN 1

WHEN l_case IN ('Tue6', 'Wed5', 'Wed6', 'Thu4', "Thu5"

, 'Thu6', 'Fri3', 'Fri4', 'Fri5', 'Fri6 '.

, "Sat2", "Sat3', 'Sat4', 'Sat5", "Sat6.

)

THEN 2

0 OTHERWISE

END

;

IF SUPERIOR (p_public_holidays) = "Y".

THEN

SELECT l_result - COUNT (*)

IN l_result

OF public_holiday p

WHERE p.calendar_day > = l_from

AND p.calendar_day<=>

AND SUBSTR (TO_CHAR (p.calendar_day

, "Dy".

'NLS_DATE_LANGUAGE = English'

)

, 1, 1 ) != 'S'

;

END IF;

RETURN l_result;

EXCEPTION

WHILE OTHERS

THEN

DBMS_OUTPUT. Put_line (' CTF nb_days (' |))

To_char (p_date_from, "MON-DD-YYYY"). ', ' ||

To_char (p_date_to, "MON-DD-YYYY"). ', ' ||

p_public_holidays | ' ) : ' || SQLERRM

);

LIFT;

END nb_days;

/

REM

REM example:

REM A) fill out the 'public_holidays '.

REM

TRUNCATE TABLE public_holiday;

INSERT INTO public_holiday VALUES (DATE ' 2012-01-01', ' new year (a Sunday)');

INSERT INTO public_holiday VALUES (DATE '' 2012-01-03, "for example");

REM (insert other days...)

COMMIT;

Call the function for some pairs of dates of REM B)

REM nb1: remove only Sat/Sun

REM nb2: also remove holidays

REM

ALTER SESSION SET nls_date_format ='Dy DD-MON-YY';

WITH some_dates AS

(SELECT DATE ' 2011-12-29'd dual FROM

UNION ALL SELECT DATE ' 2012-01-08' FROM dual

UNION ALL SELECT DATE ' 2012-01-10' FROM dual

)

SELECT d1.d 'FROM '.

, d2.d ' to THE '.

, nb_days (d1.d, d2.d, ' don't) nb1

, nb_days (d1.d, d2.d, 'Y') nb2

OF some_dates d1

some_dates d2

WHERE d1.d<=>

ORDER BY nb1, nb2, d1.d

;

GO TO NB1 NB2

------------- ------------- ---- ----

Monday, January 8, 12 Monday, January 8, 12 0 0

Friday, December 29, 11 Friday, December 29, 11 1 1

Wednesday, January 10, 12 Wednesday, January 10, 12 1 1

Monday, January 8, 12 Wednesday, January 10, 12 2 2

Friday, December 29, 11 Monday, January 8, 12 7 6

Friday, December 29, 11 Wednesday, January 10, 12 9 8

Hope that this could be useful, but note that this code has not been completely tested, so check and test before you trust it (in the case of any questions, please post a comment)

Bruno Vroman.

Best regards

Bruno

These last days on each screen by clicking on sparkles and top rollers at the bottom makes it difficult to work.

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use to insert the last day of each month
Hi, I wrote this code but it did not insert the 20 December and it is January 1, and next time running is 31 January... I would like to work to make this insert every last day of each month...
```
begin
  sys.dbms_job.submit(job => :job,
                      what => 'INSERT INTO hisrtu 
                                        SELECT rtunum, linked, dbchkd, chaqty, dbvald, ctrltr, meaadj, qtyrtu, month, year, entity, rtunam 
                                        FROM qtyrtu;',
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  commit;
end;
/
```
I don't know if it's because I assigned the date 31-10... In November it's just but December has not...

Best regards, hope you can help me.

I use oracle 9i
Hello

If you want the job to run at 02:00 the last day of the next month, then the desired argument is:
```
interval => 'ADD_MONTHS (TRUNC (SYSDATE, ''MONTH''), 2) - (22 / 24)'
```
In other words, if she finishes running at 02:30 on 31 January, then
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ADD_MONTHS (TRUNC (SYSDATE, 'MONTH'), - 1) (22 / 24) will be 22 hours earlier, i.e. 02:00 on the last day of February.

logandro wrote:
I don't know if it's because I assigned the date 31-10... In November it's just but December has not...

Do you mean this line?

next_date-online to_date (October 31, 2010 01:00:03 ',' dd-mm-yyyy hh24:mi:ss'),

I don't think that's a problem. The next_date specifies only when it work baptisms. I guess he ran on October 31; you said it was running well on 30 November, 30 days after that. Something has to happen after November 30.

Note that the interval argument is relative when the task completes. If a task takes 30 minutes to run, then
```
interval => 'SYSDATE +30'
```
means that, if it turns to
01:00:03 October 31, then it will run again about
01:30:03 November 30, around
02:00:03 on December 30.
02:30:03 January 29 and so on. If your business does that for a few seconds, it won't matter much.

If you decide to use LAST_DAY, don't forget not that the hours-minutes-seconds referring LAST_DAY are the same as those of his argument.

Download all dates for each day of a week

I need a VI that gives me the dates of the week. I use a calendar month to look at a database and fill two different graphs with the database values. A chart looks at the values of the specific day, I chose in the calendar, but the other graph looks like all the days of the week. To create this SQL query, I need to have all the dates this week.

I could build a function if I knew that the first and last date of the week. Anyone know an easy way to do this?

Use the Second to primitive time to convert your day to a cluster of time. One of the elements of the cluster is the day of the week that should allow you to easily find the end of the week (you can add 2 days and convert back or add 2 with the unit of the timestamp of origin).

Recurring appointment Set for certain days of the week each month in the calendar on Surface RT?

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Hi Lisa,

You were correct when you say that 'it seems to me that's not possible. ' The recurring appointment for some day of the week each month in the calendar on Surface RT cannot be done by date and not day.

Good day!

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When you try to run the company, I received this error for the last two days. I have tried using another browser and ended in the same results.
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Your proof of cloud creative's 'VIP' instead of the company, so you would not normally manage your CC on the Enterprise Dashboard, but you had some Digital Marketing rights so that's why you have a dashboard.

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I was working on a scenario week last on my PC at home. Yesterday, I tried to connect from my computer at work, and I was asked to reset my password before continuing. I did it according to the instructions, but when I have access to history, I found my s

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Looks like you worked in the desktop application in the history of offline through your PC at home.

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If that's true then please switch to online mode.

The product I bought doesn't work! I need help and I was stuck in your 'contact us loop' for the last few days and I'm frustrated. How can I contact you for help!

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get the last day of the week
Hi friends,

I have the below query that returns me the data as below:

Select 'Label', to_number (to_char (to_date(start_date,'yyyy-mm-dd'), 'IW')) - to_number (to_char (add_months (sysdate,-6), 'IW')) + 1 as a weekly sum, (col1) val yyy_view
where to_date(start_dt,'yyyy-mm-dd') between trunc (add_months(sysdate,-6)) and trunc (sysdate)
Group of to_char (to_date(start_date,'yyyy-mm-dd'), 'IW')
order of to_char (to_date(start_date,'yyyy-mm-dd'), 'IW')

Label weekly val
1 20 data
2 24 data
3 20 data
4 28 data
5 20 data
6 24 data

and so on...

Now the requirement is to get the last date of the week for each week as below

Label weekly val
given 2011-08-30 20
given 2011-08-23 24
given 2011-08-16 20
given 28 2011-08-09
given 2011-08-02 20

the weekly column should have the date of the last week. In the example above Wednesday is supposed to be the last date of the week.

Kind regards
Pradeep
Missed the part "Wednesday is considered date of last week. The week is Thursday to Wednesday. If so:
```
select  'Label',
         trunc(min(start_dt)) as day2,
        sum(col1 ) val
  from  yyy_view
  where start_dt between trunc(add_months(sysdate,-6)) and trunc(sysdate)
  group by trunc(start_date - 3,'IW')
  order by trunc(start_date - 3,'IW')
/
```
SY.

How to get the last day of the week?
HII

I can get the week number of calendar for a given date using

SELECT to_char (to_date('04/04/2011','MM/DD/YYYY'), 'WW') FROM dual

can any body tell me, how to get the last day of the week?

and the answer should be: 04/08/2011(8th april)
Thank you
San

Published by: sandeep9 on April 4, 2011 03:50
Perhaps this...
```
SQL> select trunc(sysdate,'WW')+6 from dual;

TRUNC(SYSDATE,'WW')+
--------------------
08-APR-2011 00:00:00

SQL>
```

First day/last day of each month
Hello
y at - it sql/script to get the first day/last day of each month.new to bi do not know the syntax

If a point on the link to get these things will be highly appreciated
Hello
I suggest that you go through the following link

(1) http://oracle.ittoolbox.com/groups/technical-functional/oracle-bi-l/obiee-date-comparison-1726677#M1729534

Toget first/last day of the month

(2) http://www.obinotes.com/2010/02/tip-to-get-firstday-lastday.html

New to OBIEE---> http://obiee101.blogspot.com/2009/07/obiee-how-to-get-started.html

Thank you
Saichand.V

last working day of each week

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