to calculate the number of days between 2 dates

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• 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.

• Dynamic calculation of the number of days between two dates in a table

  Hello

  I'm working on request where I dynamically calculate the number of days between two dates in a table.

  The calculation must be dynamic, i.e., when I recover the Start_date and End_date and move to the field following (call_duration) in the same row, the difference must be calculated dynamically in this area and make sure the field read-only.

  APEX version: 5.0

  Hi BO123,

  BO123 wrote:

  Hello

  I'm working on request where I dynamically calculate the number of days between two dates in a table.

  The calculation must be dynamic, i.e., when I recover the Start_date and End_date and move to the field following (call_duration) in the same row, the difference must be calculated dynamically in this area and make sure the field read-only.

  APEX version: 5.0

  one of the way to do this by calling ajax on change of end_date.

  See the sample code given below to fetch the resulting duration and making the field read only after calculation

  Step 1: Change your page

  under CSS-> Inline, put the code below

  .row_item_disabled {
    cursor: default;
    opacity: 0.5;
    filter: alpha(opacity=50);
    pointer-events: none;
  }
  

  Step 2: Create on demand Ajax process I say CALC_DURATION

  Please check Procces Ajax, see line 6.7 How to assign a value to the variable sent by ajax call

  Declare
    p_start_date  date;
    p_end_date    date;
    p_duration number;
  Begin
    p_start_date  := to_date(apex_application.g_x01);
    p_end_date    := to_date(apex_application.g_x02);
  
     --do your calculation and assign the output to the variable p_duration
    select p_end_date - p_start_date into p_duration
      from dual;
  
    -- return calculated duration
    sys.htp.p(p_duration);
  End;
  

  Step 3: Create the javascript function

  Change your page-> the function and the declaration of the Global Variable-> put the javascript function

  You must extract the rowid in the first place, for which you want to set the time, see line 2

  assuming f06, f07 and f08 is the id of the start date, and end date columns respectively, and duration

  See no line no 8 how set the value returned by the process of ajax at the duration column

  Replace your column to the respective column identifiers in the code below

  function f_calulate_duration(pThis) {
    var row_id  = pThis.id.substr(4);
    var start_date = $('#f06_'+row_id).val();
    apex.server.process ( "CALC_DURATION", {
    x01: start_date,x02: $(pThis).val()
  }, { success: function( pData ) {
  // set duration to duration column
  $('#f08_'+row_id).val(pData);
  // disable duration column
  $("#f08_" + row_id).attr("readonly", true).addClass('row_item_disabled'); }
  });
  }
  

  Step 4: choose the end date call the javascript function

  Go to report attributes-> edit your Date column end-> column-> Attrbiutes element attributes-> put the code below

  onchange="javascript:f_calulate_duration(this);"
  

  
  

  hope this helps you,

  Kind regards

  Jitendra
  

• JavaScript anomaly on the number of days between two dates

  Use ApEx 4.0, I found an anomaly in a javascript code that calculates the number of days between two dates, the current_date and the past_date. If the past and present is the or before March 10, 2013, and the current_date lies between 10 March 2013 and November 3, 2013, the number of days will be from 1 day to less than the actual number. Between November 3, 2013 and on 4 November 2013, the increments of number by 2, then the count will be accurate from this date forward.
  
  Here are some examples:
  
  March 10, 2013 = 69 days of December 31, 2012
  March 11, 2013 = 69 days of December 31, 2012
  March 12, 2013 = 70 days of December 31, 2012
  
  November 3, 2013 = 306 days in December 31, 2012
  November 4, 2013 = 308 days in December 31, 2012
  
  11 March should be 70 and 12 March should be 71. November 3 is 307 and 4 November corrects the number of fake, which began March 11.
  
  Change the past_date to March 10, 2013 produces the following:
  
  March 10, 2013 = 0 days of March 10, 2013
  March 11, 2013 = 0 days of March 10, 2013
  March 12, 2013 = 1 days of March 10, 2013
  
  But change the past_date to 11 March 2013, product of the correct numbers:
  
  March 11, 2013 = 0 days of March 11, 2013
  March 12, 2013 = 1 days of March 11, 2013
  March 13, 2013 = 2 days of March 11, 2013
  
  I would certainly all help to determine the cause of this anomaly. Here's the javascript code:
  
  var w1 = ($v ("P48_PAST_DATE"));
  W1 = (w1.toString ());
  vmon var = (w1.substr (3.3));
  vyr var = (w1.substr (7));
  var r = (vyr.length);
  If (r == 2)
  vyr. = (parseFloat (vyr) + 2000);
  vday var = (w1.substr (0.2));
  var y = (vmon.concat ("", vday, ",", vyr));
  y = Date.parse (y);
  
  var w2 = ($v ("P48_CURRENT_DATE"));
  var vmon2 = (w2.substr (3.3));
  var vyr2 = (w2.substr (7));
  var vday2 = (w2.substr (0.2));
  var x = (vmon2.concat ("", vday2, ",", vyr2));
  x = Date.parse (x);
  
  var numdays = (x - y);
  numdays = (Math.floor(numdays / 86400000));
  $s ("P48_NUMBEROFDAYS", numdays);

  Did you google for something like "javascript number of days between two dates. I think you will find the explanation to this observation:

  This method does not work correctly if there is an advanced economies jump between the two dates.

  There are examples available to calculate the difference between two dates.

• How to find the number of days between 2 date elements in the XSLT file

  Hello
  
  I need to calculate the number of days between 2 date elements (type xs: date). Can you please direct me as to how I can do the same thing.
  
  I work in 11g and using XSLT 1.0. I tried several options but yet to get a solution for this. I think that this can be done also using XSLT 2.0, but who has not worked for me.
  
  Can someone please help with this problem, thanks in advance!
  
  Thank you
  Anju

  Hello

  Have you seen this message?
  Re: Get the Date difference between 2 values of date in days

  You can do in the XSLT file since the dates are in ISO 8601 format.
  http://www.w3.org/TR/NOTE-datetime

  Here is a sample XSLT...

  The XSLT above will result in * 4 * for the next entry...

  
  2012-01-11T00:00:00.000-05:00
  2012-01-15T00:00:00.000-05:00
  
  

  You can test this example here...
  http://xslttest.appspot.com/

  I hope this helps...

  See you soon,.
  Vlad

  It is considered good etiquette to the answerers rewards with points (as "useful" - 5 pts - or 'correct' - 10pts)
  https://forums.Oracle.com/forums/Ann.jspa?annID=893

• 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 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.

• 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 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...

• Find the number of days between the dates...

  Hi friends,

  I am new to this development of blackberry applications so I do not know how to find the number of days between two days. Y at - there any API is avilable otherwise we have to write our own code. In fact, I tried GregorianCalendar but I get error that cannot find the symbol but I already imported net.rim.device.api.util and package java.util also... Please give an idea how to solve this problem.

  with respect,

  s.Kumaran.

  Use DateField class instances to represent dates on the screen.

  And your code will be as follows:

  long date1 = date1DateField.getDate ();

  long date2 = date2DateField.getDate ();

• Get the difference in days between 2 dates in Essbase 9.3.1

  How to calculate to get the difference in days between 2 dates in the format day of Essbase 9.3.1.la is 20090101 (yyyymmdd). If 2010907-20100917 = 10 days.
  Start and end dates may extend over several years.

  @DateDiff does not exist in the 9.3.1 I think. With this assumption...

  Could you use the CDF 'Functions of Date'? According to the docs (I've never used it):
  @JdaysOffset - returns the number of days between the firstDate and the secondDate.

  See samplecode.oracle.com for the CDF. Just search for CDF and you will find it on the second page (I think) of the FMCS.

  Or see this thread for another approach that I think the CDF under the covers: Re: Formate Date.

  Of course, you could spend to 11 for DateDiff. :)

  Kind regards

  Cameron Lackpour

• find the difference in days between two dates

  Hello world
  
  I'm trying to find out the difference in days between two dates and the execution of the query that I'm passing
  
  SELECT TO_char(sysdate, 'dd/mm/yyyy') - TO_char('15/11/2011', 'dd/mm/yyyy') DAYS FROM DUAL
  
  the error I get is
  
  ORA-01722: invalid number
  01722 00000 - "invalid number."
  * Cause:
  * Action:
  
  Could someone please help.
  
  Thanks in advance

  user10636796 wrote:
  Hello world

  Thanks a lot for all the replies. I am trying to apply it in a statement to my table like this

  SELECT trunc (sysdate) - TO_char (date_last_recommended, ' dd/mm/yyyy') DAYS OF recommendation;

  SELECT trunc (sysdate) - TRUNC (date_last_recommended) DAYS OF recommendation;

• To find the months and days between 2 dates

  Hello

  I want to find the months and days between 2 dates.

  For example.

  1 - Date: August 25, 2013

  2 - Date: October 23, 2013

  If we consider each month 30 days, it should give

  August 25, 2013 to August 30, 2013 = 6 days

  01-Sep-2013-30-Sep-2013 = 1 month

  October 23, 2013 to October 30, 2013 = 8 days

  Total = 1 month and 14 days.

  Kindly help as soon as possible.

  Thanks and greetings

  Suresh

  Assuming that d2 > d1,.

  where d)

  Select sysdate d1, sysdate + 56 double d2

  Union all select to_date (March 1, 2013 ',' dd-mon-yyyy "") d1, to_date (March 31, 2013 ',' dd-mon-yyyy ') d2 double

  Union all select to_date (5 February 2013 ',' dd-mon-yyyy ') d1, to_date (March 31, 2013 ',' dd-mon-yyyy "") double d2

  Union all select to_date (February 25, 2013 ',' dd-mon-yyyy "") d1, to_date (March 23, 2013 ',' dd-mon-yyyy ') d2 double

  Union all select to_date (February 25, 2013 ',' dd-mon-yyyy ') d1, to_date (March 31, 2013 ',' dd-mon-yyyy "") double d2

  Union all select to_date (August 2, 2013 ',' dd-mon-yyyy "") d1, to_date (29 October 2013 ',' dd-mon-yyyy ') d2 double

  Union all select to_date (February 1, 2013 ',' dd-mon-yyyy "") d1, to_date (May 31, 2013 ',' dd-mon-yyyy ') d2 double

  Union all select to_date (25 August 2013 ',' dd-mon-yyyy "") d1, to_date ('03-Sep-2013', 'Mon-dd-yyyy') d2 double

  Union all select to_date (July 30, 2013 ',' dd-mon-yyyy "") d1, to_date (August 31, 2013 ',' dd-mon-yyyy ') d2 double

  Union all select to_date (July 31, 2013 ',' dd-mon-yyyy ') d1, to_date (August 30, 2013 ',' dd-mon-yyyy "") double d2

  Union all select to_date (July 31, 2013 ',' dd-mon-yyyy ') d1, to_date (3 August 2013 ',' dd-mon-yyyy "") double d2

  Union all select to_date (3 July 2013 ',' dd-mon-yyyy "") d1, to_date (August 31, 2013 ',' dd-mon-yyyy ') d2 double

  Union all select to_date ('31-08-2013', ' dd-mm-yyyy'), to_date('05-10-2013','dd-mm-yyyy') of the double

  Union all select to_date ('05-02-2013', ' dd-mm-yyyy'), to_date('31-03-2013','dd-mm-yyyy') of the double

  Union all select to_date ('05-02-2013', ' dd-mm-yyyy'), to_date('05-03-2013','dd-mm-yyyy') of the double

  Union all select to_date ('05-02-2013', ' dd-mm-yyyy'), to_date('05-02-2013','dd-mm-yyyy') of the double

  )

  Select d1, d2,

  1 + 30 * trunc (months_between (d2, d1)) + LESS (extract (day of d2), 30)-LESS (excerpt (d1 day), 30)

  + CASE when extracted (d2 day)< extract(day="" from="" d1)="" then="" 30="" else="" 0="" end ="">

  d

  D1 D2 DAYSBETWEEN

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

  October 10, 2013 5 December 2013 56

  March 1, 2013 30 March 31, 2013

  5 February 2013 March 31, 2013 56

  February 25, 2013 March 23, 2013 29

  February 25, 2013 March 31, 2013 36

  August 2, 2013 29 October 2013 88

  February 1, 2013 may 31, 2013 120

  August 25, 2013 03 - Sep-2013 9

  July 30, 2013 31 August 31, 2013

  July 31, 2013 August 30, 2013 31

  July 31, 2013 3 August 2013 4

  July 3, 2013 August 31, 2013 58

  31 August 2013 5 October 2013 36

  5 February 2013 March 31, 2013 56

  5 February 2013 March 5, 2013 31

  February 5, 2013 February 5, 2013 1

  In my view, which corresponds to your rules.

• 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<=>