Calculate the interval in days

Similar Questions

• Calculate the number of days remaining until the next anniversary. Help, please!

  Hi guys,.

  I'm new to the forum and to the development of BB. So please do not judge harshly if the answer to my question seems obvious.

  I need to calculate the number of days until the next birthday (taking into account any valid birth date)

  After looking at the API and the forum search, I realized that I could

  calculate the difference between two dates in milliseconds and then divide the delta of the value of a day in milliseconds. That is to say:

  birthdayCalendar.set (Calendar.YEAR, those);

  Date1 = birthdayCalendar.getTime () .getTime ();

  date2 = System.currentTimeMillis ();

  Delta = date1 - date2

  numOfDays = delta\DateTimeUtilities.ONEDAY

  Here's my question. How do I create a valid date1 what about leap years?

  Maybe there's a better way to solve this problem?

  Any help is greatly appreciated.

  I agree that the determination of the number of days between today and Feb. 29 could be a little difficult, if it is not a leap year.  I suspect that the calendar has a certain built-in mechanism to compensate for this, but your solution to choose a date is better.

  DST has a role to play, and Yes, the calendar makes up for it.  Let us take two dates, for example (before DST) March 1 and June 1 (after the DST).  If you take a calendar and set the same hour, minute, etc., they are all the same except for dates, then subtract one from the other, and then divide by the number of milliseconds in 1 hour, you calculate the number of hours between the same time on two different days.  You will find that it is not a multiple of 24 hours - because there is actually an hour less than the number of days since the clocks move forward (in the northern hemisphere anyway...),

  In your case, you are not calculating the number of hours, you calculate the number of days.  But you will be dividing by [24 * (time in an hour)].  If you take [23 * (time in an hour)] and divide it by [24 * (time in an hour)], using arithmetic on integers, you will end up 0.  The easiest to get around this, who also works for the end of the DST too, is to simply add one hour to the date before the division.

  Hope that makes sense...

• 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

• How to calculate the number of days/weeks/months between 2 dates?

  Hello
  
  I would like to know how to calculate the number of days/weeks/months between 2 dates in OBIEE 11 g, for example, I have 26/05/2013 and 19/05/2013, then I want to get 7 days.
  
  Thank you!
  Jamie

  Hi Jamie,

  Through this links...

  http://www.bravesoft.com/blog/?p=682
  http://twobiee.blogspot.in/2012/01/working-with-date-differences.html

  Mark as correct it allows u...
  Thank you...

• Calculate the difference in days between two Dates

  Hello

  I'm trying to understand how to calculate the difference in days between two dates using JavaScript in LiveCycle. (Minimum = JavaScript knowledge)

  Where 'Start_Date' and 'Current_Date' are the names of the two dates in the palette of the hierarchy. (the two Date/time field)

  * Current date is using the object > value > execution property > current Date/time

  I need a text or number field showing the difference in days. (Difference_in_Days)

  I noticed the following code is pretty standard among other responses:

  var

  Start_date = new Date (Start_Date);

  var

  Current_Date = new Date (Current_Date);

  var

  nAgeMilliseconds = Current_Date.getTime) - Start_date.getTime ();

  var

  nMilliSecondsPerYear = 365 * 24 * 60 * 60 * 1000 ;

  I know there is lack of code and code above are may not be not correct.

  Please notify.

  OK, that's because of the way that javascript and works of the calculate event.  The field will be filled with whatever the script resolves at the end of execution. Technically, your script does not have a value because the last thing you do is an assignment to a variable.  Change the last line as follows:

  Math.ABS ((firstDate.getTime)

  ((- secondDate.getTime (()) / (oneDay));

  (eliminate the variable assignment) and get rid of the app.alert.  Your script will "return" (have) regardless of the value of calculation from the East and which will be stored in the field.

• to calculate the number of days between 2 dates

  Hi all
  
  
  I need to calculate the number of days between 2 dates.
  Y at - it no oracle built on that basis.
  
  
  
  Thanks in advance
  Kind regards
  
  Nick

  user646786 wrote:
  Hello

  Thanks a lot its really useful, but I must make a correction, I just want weekdays rather working days

  As already said, there are many examples on the forum of the calculation of the working days or the days of the week or whatever you want between certain dates of restrictions. You will need to have a go at adapting what has already been given before and then when you get stuck, come here and your postcode, and someone will take a look for you.

• calculate the number of days excluding non-working days

  11.2.0.4

  theres a fairly common request over the years to get the diff between 2 dates, excluding weekends, theres many ways to to do, however, I would like to go a step further and exclude everything not professional

  day.

  We have a company function that calculates whether a date is a holiday for our Organization.  We can pass a date and we have a Y or N back.     In the example below, the l' appel call function shows there for Saturday to show how we can identify the non-working day.

  Can anyone think of a way that we could use to get a sum of N s between 2 date fields in a line using this function?   IM thinking if we get the right formula/function to create a virtual column with calculating pre built as it can start to get expensive if she had to loop through every possible day between 2 dates to see it is a business day, all by a count.

  create the table t1_count

  (identification number,

  date of TS1,

  date of TS2)

  insert into t1_Count

  values (1, sysdate, sysdate - 7);

  insert into t1_Count

  values (2, sysdate, sysdate - 4);    -Saturday

  commit;

  SELECT id, ts1 ts2, .isholiday (ts1), isholiday (ts2) of t1_count;

  1	06/03/2015-14:30:51	27/05/2015 14:30:51	N	N
  1	06/03/2015-14:32:29	30/05/2015 14:32:29	N	THERE

  That fold to my brain...

  ABS (trunc (sysdate - 40) + level - 1 - trunc (sysdate)) that's equivalent to the abs (level - 41)...

  Why not just level<=>

• Calculate the number of days off weekends

  
  Hi all

  I have a requirement where I have to calculate the difference between two dates, excluding weekends and holidays. Is there a way I can do it in OBIEE analysis.

  Any help would be appreciated.

  Thanks in advance!

  Check it for weekends cool - bi.com

  For the holidays, you will need to customize you the size of the day

  check if help

• 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

• Dynamically calculate the number of days between two dates and amounts of split

  Hello
  
  I have searched for a solution for this, but had no success.
  I need to show the amounts broken down by days.
  
  I have a table that has an amount column and start and end dates.
  
  I need to write a query so that the amounts will be broken evenly based on the number of days between the start date and end date.
  
  For example, for this line.
  insert into my_test values (' 1, '' 3-mar-2010, ' 7 - mar - 2010 ", 1000);
  
  the query returns this (split $1,000 over 5 days)
  
  
  ID Date amount
  1 ' 3-mar-2010' 200,00
  1 ' 4-mar-2010' 200,00
  1 ' 5-mar-2010' 200,00
  1 ' 6-mar-2010' 200,00
  1 ' 7-mar-2010' 200,00
  
  
  
  create table my_test)
  ID number (10),
  start_date date,
  End_date date,
  amount number (10.2)
  );
  
  
  Select * from my_test
  
  insert into my_test values (' 1, '' 3-mar-2010, ' 7 - mar - 2010 ", 1000);
  insert into my_test values (2, 10-mar-2010 ", 19-mar-2010", 2000);
  insert into my_test values (3, 20-mar-2010 ',' 21-mar-2010, 5000);
  
  
  
  Thanks in advance.

  Hello

  One way is to join a Meter of Table , a table, or (more often) a set of results includes a line for eery number 1, 2, 3,... until the maximum number of times you need to divide a line.
  For example:

  WITH     cntr     AS
  (
       SELECT     LEVEL - 1     AS n
       FROM     (  SELECT  MAX (end_date - start_date)     AS max_day_cnt
               FROM        my_test
            )
       CONNECT BY     LEVEL <= 1 + max_day_cnt
  )
  SELECT       t.id
  ,       t.start_date + c.n                    AS dt
  ,       t.amount / (t.end_date + 1 - t.start_date)     AS amt
  FROM       my_test      t
  JOIN       cntr            c     ON     c.n <= t.end_date - t.start_date
  ORDER BY  id
  ,            dt
  ;
  

  This assumes that all dates have the same number of hours, minutes, and seconds, as is the case in your sample data.
  If this isn't the case, then use TRUNC (start_date) and TRUNC (end_date) instead of start_date and end_date or post some sample data and results if some lines do not represent a whole number of days.

• How to calculate the number of days between dates

  How can I determine the number of days between 2 dates? Say today and 10/07/46
  
  Thanks Bob

  You could do it like this:

  Photoperiod var = 1000 * 60 * 60 24; number of ms in a day

  var today: Date = new Date();

  var every time that: Date = new Date (1946,6,10);

  var diff = Math.floor((today.getTime()-whenever.getTime())/dayLength);

  trace (diff)

• Calculate the last business day

  Hello
  
  I use Oracle Database 11.2.1 and wants to return to the last day of work (JJ/MM/AAAA).
  
  Working days Monday to Friday delivery
  
  for example on Sunday, 18 June 2000, the returned value would be Thursday, June 15, 2000

  sliderrules wrote:
  I wanted to say the days of the week

  for example, Monday, May 13, 2013 would return 10/05/2013
  Tuesday, May 14, 2013 would return 13/05/2013 etc.

  What the earlier query?

  SQL> with t as
    2      (
    3      select to_date('13/05/2013 ','dd/mm/yyyy') dt
    4      from dual
    5      union all
    6      select to_date('14/05/2013 ','dd/mm/yyyy') dt
    7      from dual
    8      )
    9      select dt,dt -
   10             case to_char(dt,'fmday','nls_language=english')
   11                         when 'sunday' then 2
   12                 when 'monday' then 3
   13                else 1
   14            end last_w_day
   15     from t;
  
  DT        LAST_W_DA
  --------- ---------
  13-MAY-13 10-MAY-13
  14-MAY-13 13-MAY-13
  

• calculate the total of days

  Hello world
  
  I have 2 items in date:
  
  p9_begin and p9_end
  
  I want to show every Sunday between these two days, and the total number of Sundays.
  I also want to show the total number of days less on Sundays.
  
  I can't think of a solution for this. I tried to loop, but it doesn't seem to work properly.
  Does anyone have a suggestion?
  
  Kind regards
  Diana

  Hi Diana,

  You could do something like:

  WITH X AS (SELECT TO_DATE(:DATE1,'DD/MM/YYYY') + LEVEL - 1 THE_DATE FROM DUAL CONNECT BY TO_DATE(:DATE1,'DD/MM/YYYY') + LEVEL - 1 < TO_DATE(:DATE2,'DD/MM/YYYY'))
  SELECT (SELECT COUNT(*) FROM X WHERE TO_CHAR(THE_DATE, 'DY', 'NLS_DATE_LANGUAGE=ENGLISH') = 'SUN') SUNDAYS,
  (SELECT COUNT(*) FROM X WHERE TO_CHAR(THE_DATE, 'DY', 'NLS_DATE_LANGUAGE=ENGLISH') <> 'SUN') OTHERDAYS
  FROM DUAL
  

  which would give you your Sunday/non-dimanche. pus

  WITH X AS (SELECT TO_DATE(:DATE1,'DD/MM/YYYY') + LEVEL - 1 THE_DATE FROM DUAL CONNECT BY TO_DATE(:DATE1,'DD/MM/YYYY') + LEVEL - 1 < TO_DATE(:DATE2,'DD/MM/YYYY'))
  SELECT THE_DATE FROM X
  

  I will give you all the dates.

  : DATE1 is some element contains your 'count' and: DATE2 is your "to this day".

  Andy

• How to calculate the date interval

  Hi Experts,
  
  I'm looking for a way to calculate the interval between two dates. And get the result to the format "dd.mm.yyyy hh24:mi:ss.
  

  select  sysdate - to_date('1.1.2000 12:00:00','dd.mm.yyyy hh24:mi:ss') from dual

  Best practices for a problem like this?
  
  Best regards
  Igor

  select trunc( months_between( day2, day1 ) / 12 ) years
       , mod( trunc( months_between( day2, day1 ) ), 12 ) months
       , numtodsinterval( day2 - add_months( day1, trunc( months_between( day2, day1 ) ) ), 'DAY' ) rest
  from ( select to_date( '19-01-1979', 'dd-mm-yyyy' ) day1, sysdate day2 from dual )
  

• Function to calculate the days of work

  Hi all

  I accidentally develop an application in which I had to follow the thread below and it works perfectly for the weekend (Saturday and Sunday)

  PL/SQL function to calculate the non-working days

  Now, my specific question is: how to implement the same function , but weekends (Fri & sat).

  I tried to recognize the logic behind the calendar to change the days of the weekend, but I suspect.

  I need your recommendation around this function specifically.

  !-[CodeBlockStart:3cad70b5 - 81 a 0-4f6b-909 a-cacb29e25953]--> BOX WHEN IN l_case (-"Mon6", "Tue5", "Wed4", "Thu3", "Fri2" - exclude Saturday & Sunday, workday begins from Monday 'Sun6', 'Mon5', 'Tue4', 'Wed3', 'Thu2' - remove begins day 2 business days (Fri & sat) on Sunday and maintaining the order of the sequences,-"Sat1", "Sun1", "Sun2', 'Sun3', 'Sun4' - Include weekend can be for the rest of the week) , "Sat1", "Sat2", 'Fri1', "Sat3", "Sat4" - includes my week ends replace (with FRI Sat) & (Sun to SAT) & law enforcement-sequence, "Sun5', 'Sun6',"Sat5","Sat6"- replace (Sun with Sam).                                                                   ) 1 THEN WHEN IN l_case (-'Tue6', 'Wed5', 'Wed6', 'Thu4', "Thu5" - Old (Saturday and Sunday) end over the next week - I guess - sequence of days - since the week days sequence Changed (number 1 or 1 step backward) when numbers also changed accordingly "Tue5", "Wed4", "Wed5", "Thu3", "Thu4" - ok-, 'Thu6', 'Fri3', "Fri4", "Fri5", "Fri6", "Thu5", "Sun1", "Sun2", 'Sun3', 'Sun4' - contiue to Thu sequence number & then replace my current (Friday) holiday with start of work dat (Sun) "Fri1") (, 'Fri2', 'Fri3', 'Fri4', "Fri5") THEN 2 0 OTHERWISE END;

  Thanks in advance.

  Amatu Allah

  with

  calendar as

  (select dat_val,

  cases where to_char(dat_val,'mm-dd') in (12-25', 12-26', ' 01-01')-in fact select the host table to_char(holiday_date,'mm-dd')

  then "vacation."

  When to_char (dat_val, 'Dy') in ('Friday', 'Saturday')

  then 'weekend '.

  else 'work '.

  end dat_type

  sum (case when to_char(dat_val,'mm-dd') in (12-25', 12-26', ' 01-01'))

  then 0

  When to_char (dat_val, 'Dy') in ('Friday', 'Saturday')

  then 0

  1 other

  end

  ) on working_count (dat_val control)

  from (select date "' 2015-12-01 + level - 1 dat_val)

  of the double

  connect by level<=>

  )

  )

  Select dat_val,

  dat_type,

  working_count

  calendar

  where dat_val between to_date(:lo_date,'yyyymmdd') and to_date(:hi_date,'yyyymmdd')

  and dat_type! = work «»

  DAT_VAL	DAT_TYPE	WORKING_COUNT
  11/12/2015	end of week	8
  12/12/2015	end of week	8
  18/12/2015	end of week	13
  19/12/2015	end of week	13
  25/12/2015	holiday	18
  26/12/2015	holiday	18
  01/01/2016	holiday	23
  02/01/2016	end of week	23
  08/01/2016	end of week	28
  09/01/2016	end of week	28

  Select to_date(:lo_date,'yyyymmdd') lo_date,

  to_date(:hi_date,'yyyymmdd') hi_date,

  dat_val,

  dat_type,

  working_count,

  Lead (working_count) more (order of dat_val) - working_count + 1 working_days,

  to_date(:hi_date,'yyyymmdd') - to_date(:lo_date,'yyyymmdd') - lead (working_count) (order dat_val) + working_count non_working_days

  calendar

  where dat_val in (to_date (:hi_date,'yyyymmdd'),to_date(:lo_date,'yyyymmdd'))

  LO_DATE	HI_DATE	DAT_VAL	DAT_TYPE	WORKING_COUNT	WORKING_DAYS	NON_WORKING_DAYS
  10/12/2015	10/01/2016	10/12/2015	work	8	22	10
  10/12/2015	10/01/2016	10/01/2016	work	29	-	-

  Concerning

  Etbin