Good count of days between two dates

Similar Questions

• counting the days between two dates

  I have 3 columns in my table
  
  date of purchase date sold days
  computer 24 April 09-11 May 09
  
  
  How can I write that a sql statement will help me to create a new column days and days between date bought and sold
  
  
  Thank you

  872959 wrote:
  I have 3 columns in my table

  date of purchase date sold days
  computer 24 April 09-11 May 09

  How can I write that a sql statement will help me to create a new column days and days between date bought and sold

  Thank you

  just subtract two dates & the difference will be in units of days

  SQL > select to_date('11-MAY-09','DD-MON-YY') - double TO_DATE('24-APR-09','DD-MON-YY');

  TO_DATE('11-MAY-09','DD-MON-YY')-TO_DATE('24-APR-09','DD-MON-YY')
  -----------------------------------------------------------------
  17

  Published by: sb92075 on July 19, 2011 10:26

• 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.

• 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
  

• Days between two dates given

  Hi all

  I need the days between two dates given.

  Select to_date('25-mar-2014','dd-mon-yyyy') start_date, end_date to_date('30-mar-2014','dd-mon-yyyy')

  of the double

  Desired output:

  25

  26

  27

  28

  29

  30

  Thanks in advance.

  Kind regards

  Afzal.

  Select the level - 1 desired_output + to_number (to_char (start_date, 'dd'))

  of (to_date('25-mar-2014','dd-mon-yyyy') select start_date, end_date to_date('30-mar-2014','dd-mon-yyyy'))

  of the double

  )

  connect by level<=  end_date="" -="" start_date="" +="">

  DESIRED_OUTPUT
  25
  26
  27
  28
  29
  30

  Concerning

  Etbin

• How to find days between two dates in days

  Hi, I want to get the days between two dates, expressed in days, how can I do this?  For example, I date i.e. 1 October 10, 2013 (Thursday) and 2 as October 13, 2013 (Sunday).  How can I get a production of:

  Thursday, Friday, Saturday and Sunday, which are the days between Oct. 10-13, 2013.  I hope you understand my point.  Thank you

  You can try something like this

  CURSOR LOAD_PROF is

  SELECT load_prof1, v_time_profile_day

  OF load_profile_test

  WHERE profile_day IN (select TO_CHAR(:day1+level-1,'Day')

  of the double

  connect by level<>

  );

  Kind regards

  S Pax

• 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;

• 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.

• 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

• Behavior of days between two dates

  Hello!
  If anyone can help? ... with differences in Date? :
  
  All I want is to calculate the number of days between two dates:
  
  InitDate has been entered as 01/10/2010 planning, to recover is 20101001
  HireDate was entered into the 1980-10-01 planning, to recover is 19801001
  
  I tried
  
  NumDays = @DATEDIFF (InitDate, HireDate, DP_DAY);
  
  but give me only 0 and -1.
  
  
  
  When I try the regular difference:
  
  NumDays = InitDate - HireDate;
  
  He returned to 30,000 (which in the example is 20101001-19801001) - which should not.
  
  
  I use Essbase/Planning 11.1.1.2, I tried also to break dates in years and months using INT and operations, but when I want to get the calendar days he is ugly. The two dates live in the same block.
  
  Someone at - it an idea?
  
  Thank you very much!

  Hello
  Here's a piece of code in my trunk of code which may help you. This assumes you have string CDF pack (https://www.samplecode.oracle.com/tracker/tracking/linkid/prpl1004/remcurreport/true/template/ViewIssue.vm/id/S441/nbrresults/9)

  var y1, y2, m1, m2, d1, d2, difx1, difx2;

  FIX (SOLUTION YOUR MEMBERS HERE)
  "BLOCK MEMBER")

  Y1=@round ("Date of 1" / 10000,0);
  Y2=@round ("Date 2" / 10000,0);
  M1=@round (("Date 1"-y1*10000)/100, 0);
  M2=@round (("Date 2"-y2*10000)/100, 0);
  D1 = ' Date of 1 "-(y1*10000+m1*100);"»
  D2 = "Date 2"-(y2*10000+m2*100); "»

  difx1=@TODATEEX ("mm/dd/yyyy",
  @JconcatStrings (@LIST (@JgetStringFromDouble (m2,@_false,@_false), ' / ', @JgetStringFromDouble (d2,@_false,@_false), ' / ', @JgetStringFromDouble (y2,@_false,@_false)))
  )
  ) ;
  difx2 = @TODATEEX ("mm/dd/yyyy", @JconcatStrings (@LIST)
  @JgetStringFromDouble (m1,@_false,@_false), ' / '.
  @JgetStringFromDouble (d1,@_false,@_false), ' / '.
  @JgetStringFromDouble (y1,@_false,@_false))
  )
  );

  diff=@DateDiff (difx2, difx1, DP_DAY);
  )
  ENDFIX

  See you soon,.
  Alp

• County of days between two dates without weekend

  Hello
  I need a solution in the query or another thread, which returns the number of days between two dates without considering the weekend (Saturday and Sunday), I have a column of type Date, and return need in the form of a column HH hours and days in another column.
  
  Concerning
  Jonas

  Hi and welcome to the forum.

  Don't forget you can do a search on this forum.
  Your question has been asked before.

  Some other tips:

  http://asktom.Oracle.com/pls/asktom/asktom.download_file?p_file=6551242712657900129
  http://asktom.Oracle.com/pls/asktom/f?p=100:11:0:P11_QUESTION_ID:185012348071

• Number of occurrences of a day between two dates

  Hello

  I have a requirement in my code to count the number of times per day occurs between two dates. So, for example if

  start date = January 1, 2014

  end date = March 31, 2014

  The number of times that happens day 1 is three times. That's why I want the result to be 3. My three input parameters are the number of the day, start and end dates. I know that this can be done by parsing through the days and incrementing a counter of sorts, but can it be done without any loop etc?

  Looking for options of

  Thank you

  Birdy

  SELECT COUNT (TO_CHAR (DAYS, 'DD')) IN)

  SELECT DAYS START_DATE + LEVEL 1

  FROM T1

  CONNECT BY LEVEL<=>

  WHERE TO_CHAR (DAYS, 'DD') ='&01';

  Post edited by: Parth272025

• days between two dates, excluding Saturday and Sunday

  Hello
  I have an obligation as: there are two entries in the query 1) startdate and 2) end date when the end user enter the start date and end date of the sql query to retrieve the number of days between given dates of beginning and end, and it should be retrivied with two condition 1) every Sunday between the above dates excluded 2) second and fourth Saturday of the month between the start dates and should be excluded.
  example of start date: 01-May-09
  end date: 16-May-09
  
  release date: 13

  SELECT COUNT (*)
  (SELECT TO_DATE('01-may-09') + LEVEL - dates 1)
  OF THE DOUBLE
  CONNECT BY LEVEL<= (to_date('16-may-09')="" -="" to_date('01-may-09'))="" +="" 1)="">
  WHERE ((TO_CHAR (inner.dates, 'd') = 7
  AND TO_CHAR(inner.dates, 'W') NOT IN (2, 4)
  )
  GOLD TO_CHAR (inner.dates, 'd.') not in (1.7)
  );

• 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.

• Figure out how many years and days between two dates

  Nice day

  I've been dealing with the problem for a few days now so I thought that I could send it and see if someone can tell me what I'm doing wrong.

  / * Create a Date with a date object. */

  var d1 = getField "(RELEASE. DATE') .value;

  / * Create a date object containing the current date. */

  var d2 = getField("Text4").value;

  / * Number of years difference. */

  diff = (((d2.valueOf () - d1.valueOf (()) / 1000) / 60) / 60) / 24) / 365;

  Displays the field "Text3".

  getField("Text3").value = (diff); & #8232;
  < END >

  RELEASE. Value DATE is January 1, 2011, with a date format dd/mmm/yyyy

  Text4 value is March 1, 2011 with a date format dd/mmm/yyyy

  The release of Text3 gives me a 'NaN' result

  What I'm trying to do is calculated the number of years an employee has had with the company, and then display a balance of days that he had.

  Any thoughts would be greatly appreciated.

  repeated conversion from date string and number of days for the reuse of code
  function Date2Num (cFormat, cDate) {}
  Converts a date string formatted in days since the date of the time
  Kai var = util.scand (cFormat, cDate);
  Math.floor (oDate.getTime (return) / (1000 * 60 * 60 * 24));
  }

  / * Create a Date with a date object. */
  var d1 = (getField("Enrol_Day").value + "/" + getField("EnrolMonth").value + "/" + getField("EnrolYear").value);
  Converts a date in the date object string
  oD1 var = util.scand ("dd/mmm/yyyy", d1);
  get the current date of the field
  var d2 = (getField ("TD. RelDay") .value + ' / ' + getField ("TD. RelMonth") .value + ' / ' + getField ("TD. (RelYear') .value);

  calculate the difference in days
  diff var = Date2Num ("dd/mmm/yyyy", d2) - Date2Num ("dd/mmm/yyyy", d1);
  increment the value if the end date should be included in the calculation
  diff ++;
  convert days to truncate all years - to the next intiger lowest
  var carnavalSVP = Math.floor(diff / 365.25);
  get the current year interval - rest of diff days divided by 365.25 days
  var nDays = Math.floor (diff % 365.25);

  display the entered values
  CMSG var = ' end date: "+ d2 +" \t Compute days: ' "+ Date2Num (" dd/mmm/yyyy", d2);
  CMSG += "\nStart date:" + d1 + "\t Compute days: '" + Date2Num ("dd/mmm/yyyy", d1);
  Total difference in days
  CMSG += "\nTotal elapsed days:"+ diff;»
  result in years and days
  CMSG += "\nYears:" + carnavalSVP + "and" + nDays + 'days'; "."
  App.Alert (GSMC, 1, 0, "results");
  //