Function to calculate the days of work

Similar Questions

• Calculation of the days of working in a function

  I have a function that will count the total number of days between two weeks. That is, May 19, 2010 to June 18, 2010 is 30 total days in my cycle. We do not work on weekend so that 30 days is not correct. The formula below is how I calculate the days between cycles. Is it possible to not counting the weekend in my cycle? Please see the formula below:
  
  TIMESTAMPDIFF (SQL_TSI_DAY, "Free Dim". "Release cycle Start Date ', 'Release Sun'." Output cycle end Date")

  You should subtract the start end date date, divide by 7 and multiply by 5.

  Try with it.

  Concerning

• Calculate the days of salary

  Dear friend
  
  
  I want to calculate the days of salary. We pay the salaries of the 21st of the month last to the 20th of the current month.
  
  I want to create a view where it will show of the the Scriptures based on this month wise condition
  
  Select empno, name, daytype, count (empno) days, MAX (TO_CHAR (ADATE, 'MONTH')) in dailyattendence
  When al > =('20-OCT-2010') AND ADATE < =('20-NOV-2010')
  Group by empno, daytype
  order by empno
  
  
  for ex
  
  562 SEUN BÉLANGER AFTERNOON OCTOBER 1
  562 SEUN BÉLANGER MORNING OCTOBER 2
  562 SEUN BÉLANGER NIGHT OCTOBER 2
  562 SEUN OFF BÉLANGER 1 OCTOBER
  
  562 SEUN BÉLANGER AFTERNOON NOVEMBER 1
  562 SEUN BÉLANGER MORNING 2 NOVEMBER
  562 SEUN BÉLANGER NIGHT NOVEMBER 2
  562 SEUN OFF BÉLANGER 1 NOVEMBER
  
  Suggest me how to set the dates so that I can get the report as shown above,
  
  Sandy

  It doesn't really have anything to do with the forms, but here's a solution.
  To get the 21st of the month before you use:

  select to_date('21'||to_char(add_months(sysdate,-1),'mmyyyy'),'ddmmyyyy')
  from dual;
  
  TO_DATE('
  ---------
  21-OCT-10
  

  The same for the 20th of this month. So you get:

  where adate between to_date('21'||to_char(add_months(sysdate,-1),'mmyyyy'),'ddmmyyyy')
     and to_date('20'||to_char(sysdate,'mmyyyy'),'ddmmyyyy')
  

  An addition: If your date includes a time, you must use the use trunc (date).

  Published by: InoL on November 11, 2010 12:10

• 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

• Function to calculate the Minutes of business between two Dates

  Hello

  I use the version of database Oracle 11.2.1. I would like to calculate the number of minutes of work between the two dates where working hours are Monday - Thursday from 08:45 to 17:00 and Friday from 08:45 to 16:30

  I found the following function useful in a similar position but wanted help picking up the days and the hours above into account

  CREATE OR REPLACE
  FUNCTION get_bus_minutes_between(
                                   p_start_date DATE,
                                   p_end_date DATE
                                  )
    RETURN NUMBER
    IS
        v_return NUMBER;
    BEGIN
        with t as (
                   select  case level
                             when 1 then greatest(p_start_date,trunc(p_start_date) + 9 / 24)
                             else trunc(p_start_date) + level - 15 / 24
                           end start_dt,
                           case connect_by_isleaf
                             when 1 then least(p_end_date,trunc(p_end_date) + 18 / 24)
                             else trunc(p_start_date) + level - 8 / 24
                           end end_dt
                     from  dual
                     connect by level <= trunc(p_end_date) - trunc(p_start_date) + 1
                  )
        select  sum(greatest(end_dt - start_dt,0)) * 24 * 60 work_minutes
          into  v_return
          from  t
          where trunc(start_dt) - trunc(start_dt,'iw') < 5; -- exclude weekends
        RETURN v_return;
END;
/

  Can anyone help?

  Hello

  Think about what you do.

  sliderrules wrote:

  SELECT TO_CHAR (DATE ' 1-1-1' + business_end_time, 'HH24:MI:SS')-TO_CHAR (DATE ' 1-1-1' + business_start_time, 'HH24:MI:SS') * 24 * 60

  of time_dim

  where full_date = July 1, 2013.

  AND working_day = 'Y ';

  The following error is returned:

  ORA-01722: invalid number

  The time is converted to character reslting with this error. Is there a better way to subrtacting types of data INTERVAL DAY (1) SECOND (2) (business_end_time - business_start_time)

  To_char, as its name implies, returns a string of characters.  You say that

  TO_CHAR (...) - TO_CHAR (...) * 24 * 60

  If you ask for the string returned by the 2nd TO_CHAR be multiplied by 24.  Does make any sense to you?

  Maybe you wanted to do the subtraction first, which would be

  (TO_CHAR (...) - TO_CHAR (...)) * 24 * 60

  but still would be a mistake, because you can't subtract a string to another string (or anything else).

  .

  Last Friday, I proposed this:

  SELECT SUM (LESS (end_dt, v_end_dt)

  -Most GREAT (start_dt, v_start_dt)

  ) * 24 * 60...

  where start_dt and end_dt were the DATE in the table columns, as the value returned by TO_DATE (08 11 July 2013:45 ', "DD-Mon-YYYY HH24").

  If I understand correctly, you would store stores rather 3 columns: full_date (a DATE, where the time part is 00:00:00) as well as business_start_time and business_end_time (the two INTERVAL DAY TO SECOND).  In this case, you can modify the code above to say:

  SELECT SUM (LESS (full_date + business_end_time, v_end_dt)

  -GREATEST (full_date + business_start_time, v_start_dt)

  ) * 24 * 60...

  You can add a DATE at an INTERVAL;  the result will be a DATE.  There is no need to convert anything to a data type to the other.

• How to create a function to calculate the calendar

  Oracle Forms 6i
  
  Hai
  
  I'm generating a daily presence that is stored in the database with the fields
  
  Empcode, attend_date, intimate, outtime, extratime, deptcode, unitid and some other things
  
  I need to create a function by taking these fields of the table unitid, empcode, intimate, outtime, attend_date, deptcode and return total hours he worked, and more than 8 means it will add in extra time.
  
  If it is possible to give the run of the date value to which to this day, you must calculate.
  
  Thanks in advance
  
  Srikkanth.Ml

  Srikkanth,
  Depending on how you stated your RESPONDENT and OUTTIME, columns this should quite simply be a calculation. For example, if you have declared these columns as DATE columns, the following calculation should work:

  SELECT NVL(OUTTIME,SYSDATE) - INTIME AS hours_worked
  FROM DAILY_ATTENDANCE
  WHERE EMPCODE = "Emp Code of the employee you are evaluating";
  

  Once you have the HOURS_WORKED, you can check if the amount is greater than 8 (or whatever is the approved duration of working hours of the employee's) to determine if he or she worked longer than their approved shift back the extra time, and the user has worked or NULL or zero if they have not exceeded their shift.

  If you have set up the columns of time as something other than the DATE or DATETIME, you will need to convert the values in input/OUTPUT at a DATE so that you can use the functionality of calculation date default Oracle or you need to build your own process of calculation of the WORK_HOURS based on how you store the RESPONDENT and OUTTIME values.

  I hope this helps.
  Craig...

• calculate the day of the year using the date.

  Hi all
  Can calculate us the day of the year using the date form? Is this possible? Please help me thanks in advance
  Like this
  30/08/2009-30/08/2010 after calculation - 365
  
  
  Sarah
  
  Published by: SarahSarahSarah on August 30, 2009 12:49 AM

  Sarah.

  : ins1.noday must be a part of number!

  Create a function in forms like this:

  function get_nodays ( f_dtm_start in date,
                        f_dtm_end   in date ) return number is
  begin
  if
    f_dtm_start is null OR
    f_dtm_end is null
  then
    return ( null );
  end if;
  return ( trunc ( f_dtm_start ) - trunc ( f_dtm_end ) );
  exception when others
  then message ( sqlerrm || ' in function get_nodays occured.' );
  return ( null );
  end;
  

  Call this function in the when-validate-point triggers on ins1. INSPERIOD and: ins1.end love

  :ins1.noday := get_nodays ( :ins1.INSPERIOD, :ins1.end );
  

  It is redundant to have the: noday as a cause of part of database
  the value may by calculated every time using the values INSPERIOD and: end.

• I get "HTTP 400 error. The request is malformed. "When you try to access a URL that I consulted earlier in the day and works OK with IE.

  Hello
  I accessed the site earlier in the day, but when I came back to it I got the above message. Tried to access the site in IE and all worked well. Tried to disable DNS prefetch and erase history, but nothing helped. I did a virus scan but no threat detected.

  This problem may be caused by corrupted cookies.

  • http://KB.mozillazine.org/cookies

  Clear the cache and cookies from sites that cause problems.

  "Clear the Cache":

  • Tools > Options > advanced > network > storage (Cache) offline: 'clear now '.

  'Delete Cookies' sites causing problems:

  • Tools > Options > privacy > Cookies: "show the Cookies".

• to calculate the days of sheets each month

  Hi all

  I need a query which can get leaves from an employee of days each month.

  Example:

  1.

  If an employee has leave of date_start = 15 January 2014 ', date_end = 30 January 2014 "which is in the same month, then it will be"

  Duration = 15-30 + 1 = 16 days

  Total_days_in_jan = 31 days

  Calculation of salary leave will be amount = (: base +: housing) * (: duration) / (Total_days_in_jan)

  = (1000 + 2000) * 16/31

  ' 2. If his two month date_start = 15 January 2014 ', date_end = 10 February 2014.

  I wrote code that works perfect below

  If to_char(:date_start,'mm') = to_char(:date_end,'mm') then

  v_mon_days: = TO_CHAR (last_day (TO_DATE(:date_start,'DD-MON-YYYY')), 'DD');

  v_tot: = (nvl(:BASIC1,0) + nvl (:FUJ_LIVING,0)) *(:Duration_)/v_mon_days;

  elsif to_char(:date_start,'mm') <>to_char(:date_end,'mm') then

  v_mon_days_st_dt: = TO_CHAR (last_day(:date_start), 'DD');

  v_mon_days_en_dt: = TO_CHAR (last_day(:date_end), 'DD');

  v_lv_days_st_dt: = last_day(:date_start) -: date_start + 1;

  v_lv_days_en_dt: =: date_end - trunc(:date_end,'mm') + 1;

  v_tot_st_dt: = (nvl(:BASIC1,0) + nvl(:FUJ_LIVING,0)) * (v_lv_days_st_dt) / v_mon_days_st_dt;

  v_tot_en_dt: = (nvl(:BASIC1,0) + nvl(:FUJ_LIVING,0)) * (v_lv_days_en_dt) / v_mon_days_en_dt; 

  v_tot: = nvl(v_tot_st_dt,0) + nvl(v_tot_en_dt,0);

  end if;

  Return round (v_tot);

  3. If the 3 months off, say date_start = 24-may-2014 date_end = July 2, 2014.

  I need some help here... what happens if the 4 month also leave...

  Please suggest.

  Kind regards

  Afzal.

  Hi Roger, Etbin

  Thanks for your replies.

  I found the solution on the month to display between two dates.

  To calculate that I used the loop as below

  declare

  ST DATE: =: date_start;--TO_DATE('24-MAY-2014','DD-MON-YYYY');

  ED DATE: =: date_end;--TO_DATE('02-JUL-2014','DD-MON-YYYY');

  number of v_t_amount;

  CURSOR C1 IS

  SELECT more GRAND (ST, ADD_MONTHS (TRUNC (st, 'MONTH'), (LEVEL - 1))) dt.

  LESS (ed,

  LAST_DAY (ADD_MONTHS (TRUNC (st, 'MONTH'), (LEVEL - 1)))

  ) ld

  OF THE DOUBLE

  CONNECT BY LEVEL<>

  MONTHS_BETWEEN (TRUNC (ed, 'MONTH'), TRUNC (st, 'MONTH'))

  + 1

  ORDER BY 1;

  V_AMOUNT NUMBER: = 0;

  V_T_AMOUNT NUMBER: = 0;

  NUMBER OF V_DAYS;

  BEGIN

  FOR I IN C1 LOOP

  V_AMOUNT: = (nvl(:SALARY_DIFF,0) + nvl(:SPECIAL_ALLOWANCE,0) + nvl(:CF_OTHERS,0)

  + NVL(:Computer,0) + nvl(:PHONE_ALLOW,0) + nvl(:WORK_NATURAL,0)

  (+ nvl(:TRANSPORTATION,0) * (i.1d - i.dt + 1) / TO_CHAR (LAST_DAY (i.1d), 'dd'));

  V_T_AMOUNT: = V_AMOUNT;

  end loop;

  end if;

  dbms_output.put_line (v_t_amount);

  end;

• HP-15-af114AU: control keys for brightness (function keys) on the keyboard does work more (f2 and f3)

  Key (F2 and F3) brightness has stopped working suddenly

  I use the laptop HP 15-af114AU (AMD A8 Processor). Running on Windows 10-64 bit

  Its been only 2 weeks, I bought. The brightness of key i, e, f2 and f3 were working, then after a few days, I noticed it wasn't.
  I tried to do it manually through the brightness on the settings option, although there was no change.
  From the moment I bought this laptop, I downloaded SKYPE, Chrome and some THEMES for google Chrome.
  All other funtion keys work except these.
  Please help me with this problem.

  Hello

  Try the following.

  Download the AMD HD graphical installer on the following link and save it in your downloads folder.

  http://ftp.HP.com/pub/SoftPaq/sp74501-75000/sp74716.exe

  One when you're done, open your download folder, click with the right button on the setup of the AMD graphics card and select "Run as Administrator" to launch the installation.

  When this is finished, stop the laptop.

  Disconnect the power adapter and remove the battery.  Press and hold the Power button for 30 seconds.  Reinsert the battery and plug in the AC adapter.

  Robinet tap away at the key esc that you start the computer to launch the Start Menu, and then select f10 to open the bios menu.  Press f9 to load the default values (sometimes it is f5, but the menu below indicates the correct key), use the arrow keys to select 'Yes' and press ENTER.  Press f10 to save the setting and still use the arrow keys to select 'Yes' and press on enter.

  Start Notepad and let Windows take over completely for a few minutes before checking.

  Kind regards

  DP - K

• calculate the days remaining until a given date

  Hello

  I am facing a problem and can not find a good solution.

  I want to calculate how many days there are still up to a certain date.

  It is because I want to build in a callback to the re-kalibrate hardware which is used every year.

  So if it's today 05/11/2012, and the constant is 21/05/2012 labview must calculate there are still 10 days left before the re-kalibration must be finished.

  Someone has any idea how to do this?

  Thanks in advantage!

  of course, don't forget that time stamps are DBL in seconds.

  

  Hope this helps

• How can I use the SUM function to calculate the number of employees in the comp. &amp; deptnt?

  I have two tables: employees and their departments. I'm figuring the total employees by the Department and the total employees of the entire society. I know I have to use the SUM function, but I can only calculate total employees by Department and company separately. I need to get this result:
  
  

  DEPT_NAME     DEPT_TOTAL_SALARY         COMPANY_TOTAL_SALARY
  RESEARCH                  10875                        29025
  SALES                      9400                        29025    
  ACCOUNTING                 8750                        29025      
  
  
  This is my code:
  
  SELECT department_name, SUM(salary) as total_salary
  FROM employee, department
  WHERE employee.department_id = department.department_id
  GROUP BY department_name;
  
  SELECT SUM(salary)
  FROM employee;
  
  
  
  
  
  
  Can somebody help please?
  Thank you in advance.

  Published by: user13675672 on January 30, 2011 14:29
  
  Published by: user13675672 on January 30, 2011 14:31

  Hello

  Something like:

  SELECT dname,
         dept_tot_sal,
         SUM (dept_tot_sal) OVER () comp_tot_sal
  FROM   (SELECT   dname,
                   SUM (sal) dept_tot_sal
          FROM     dept, emp
          WHERE    dept.deptno = emp.deptno
          GROUP BY dname);
  

  There might be a smarter way, with no re - select.

  Concerning
  Peter

  Analytical functions:
  http://download.Oracle.com/docs/CD/E11882_01/server.112/e17118/functions004.htm

• How to calculate the days with 2 Date/time fields

  Hello
  I tried to create a workflow to update a numeric field with the following formula, however, I get an error message "value too long for the field 'ZNum_0' (maximum size 16) (SBL-DAT-00235). Can I know what is the problem with my calculation?
  
  dtCloseDate - dtAssignDate
  
  Thank you

  Hi, your formula is correct.

  You could be facing this problem because it will assess the value in decimal and result has no set limit for the decimal point. As digital field accept only 16 characters, the problem is occure.

  You must truncate the result with a defined decimal.

  Hope this will help you
  Dinesh

• Calculate the number of days between two dates

  Hello

  Can someone help please change my formcalc script to calculate the number of working days between two date fields.  My script currently calculates the total number of days between two dates, including the weekends which must be excluded from the total.

  If

  (HasValue (Start_Date1) & HasValue (End_Date1)) then

  $

  = Date2Num (End_Date1, "YYYY-MM-DD" "en_IE") - Date2Num (Start_Date1, "YYYY-MM-DD" "en_IE") + 1

  on the other

  ""

  endif

  Any help will be most appreciated.

  Thank you.

  Check...

  (1) you said that you put the script on the event «days1» calculate My sample imitates the variable names used in the original message, "Start_Date1" and "End_Date1". If the names of variables for the start and end dates are different, you will need to modify the script to account for these names.

  (2) the Date2Num functions in the calculation of the "totalDays" use the date format "YYYY-MM-DD". If your date habits differ from "YYYY-MM-DD" FormCalc will complain.

  Steve

• Is there a function to get the name of the days?

  Hello
  I use 10.2.0.1.0 oracle
  Is there a function to obtain the day of the week?
  as I need to get Sunday, Monday, Tuesday, Saturday.
  I know that I can use union and select the name from a double.
  But I just want to know if there is another way.
  I need to show 7 days in a poplist in my form, it's the requirement
  
  Thank you

  SELECT TO_CHAR(TRUNC(SYSDATE ,'DAY')+Level-1,'DAY') DAYS FROM DUAL connect by Level<8
  SQL> /
  
  DAYS
  ---------
  SUNDAY
  MONDAY
  TUESDAY
  WEDNESDAY
  THURSDAY
  FRIDAY
  SATURDAY