How to find the difference "within weeks" between two date values?

Similar Questions

• 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.

• How to make the difference in percentage between two matrices?

  Hello everyone, I just want to ask how to make the difference in percentage between two binary matrices. This is to compare two images converted to matrices in LabView. Thank you.

  Well, in the current image interesting pixels are zero in the reference image. Here is a small example that would take all the zero 100% pixel, then look at the pixels of same in the other image to see how are always 0.

  Modify if needed.

  

• 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

• Calculation of the difference in days between two Dates

  Hi all
  I have two text fields with dates in them (P2_CURRENT_TIME_POINT_TEST_START_DATE, P2_ORGINAL_TEST_START_DATE). I would like to assign another text field (P3_TIME_FROM_INITIAL_TEST), the difference between the two days. I use a calculation page and tried suptracting them right or a date function diff, I found online, but nothing seems to work. Could someone lend me a hand on this one?
  
  Thank you.

  Ben,

  Sorry, please try the following:

  :P3_TIME_FROM_INITIAL_TEST := TO_DATE(:P2_CURRENT_TIME_POINT_TEST_START_DATE, 'DD-MON-YYYY') - TO_DATE(:P2_ORGINAL_TEST_START_DATE, 'DD-MON-YYYY');
  

  Remember to substitute the correct format string MON-DD-YYYY.

  Kind regards
  Dan

  http://danielmcghan.us
  http://sourceforge.NET/projects/tapigen

  You can reward this answer by marking as being useful or correct ;-)

• Number of weeks between two dates that are in the same or different year

  Hello

  How to calculate the number of weeks between two dates?

  for example. "" "17-mar-2013' and June 27, 2013" or June 15, 2013 "and" 25-mar-2014
  

  
  

  Thanks in advance.

  lukx
  

  Hello

  Looks like your date range is incorrect.

  Please try the below.

  SQL >-front right

  SQL > select D2 - D1 "days."

  2 round ((D2-D1)/7.0) "weeks."

  3 round (Months_between (d1, d2), 0) 'months '.

  (select 4

  5 TO_DATE ('20121201', 'YYYYMMDD') D1, D2 TO_DATE ('20130116', 'YYYYMMDD')

  6 from two

  (7) date_range;

  Elapsed days weeks months

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

  46         7         1

• How to find the difference between standard edition and standard edition one

  How to find the difference between oracle database standard edition and standard edition one 64 bit
  
  (a) using sql
  (b) using the configuration/installation files

  How to find the difference between oracle database standard edition and standard edition one 64 bit

  (a) using sql

  Select * the option of $ v;

  (b) using the configuration/installation files

  opatch lsinventory-details

• 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

• 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.

• 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 difference between two dates in time except Sunday

  Hi all
  
  I have a table, as shown below.

  SQL> select * from test;
  
  TR_ID                                              CREATE_TIME                                                                       CODE
  -------------------------------------------------- --------------------------------------------------------------------------- ----------
  S12341                                             05-JUN-12 12.20.52.403000 AM                                                      1003
  S12342                                             11-JUN-12 11.15.33.182000 AM                                                      1003
  S12342                                             07-JUN-12 12.00.36.573000 PM                                                      1002
  S12343                                             20-JUN-12 12.34.37.102000 AM                                                      1003
  S12343                                             15-JUN-12 11.34.27.141000 PM                                                      1002
  S12344                                             01-JUL-12 10.01.06.657000 PM                                                      1002
  S12344                                             06-JUL-12 12.01.04.188000 AM                                                      1003
  S12341                                             31-MAY-12 11.20.38.529000 PM                                                      1002

  I would like to know the difference between same tr_ids create_time, which should give out in hours except Sunday.
  
  For example:
  
  TR_ID: S12344
  1002_Create_time: July 1, 12 PM 10.01.06.657000 (i.e. Sunday)
  1003_Create_time: 12.01.04.188000 AM 6 July 12
  
  1002 create time is 22:00 Sunday.
  
  If the query must exclude only the hours of Sunday which is 10 p.m. to Monday 00 h which is 2 Hrs.
  
  I tried the sub query after doing a search on this forum but I am not getting the desired output.
  

  SELECT count(*) FROM (SELECT ROWNUM RNUM,tr_id,create_time CT_1002 FROM test c WHERE c.tr_id='S12344' and 
  ROWNUM <= (select (cast(a.create_time as date)-cast((select create_time from test b where a.tr_id=b.tr_id and code=1002) as date)) 
  from test a where a.code=1003 and a.tr_id=c.tr_id) + 1) d 
  WHERE to_char(cast((select create_time from test e where e.tr_id=d.tr_id and code=1002) as date) + RNUM - 1, 'DY') NOT IN('SUN');

  Need help to get the desired o/p

  Hello

  If I unederstand the problem correctly, that's what you want:

  WITH       got_extrema     AS
  (
       SELECT       tr_id
       ,       CAST (MIN (create_time) AS DATE)     AS start_date
       ,       CAST (MAX (create_time) AS DATE)     AS end_date
       FROM       test
       GROUP BY  tr_id
  )
  SELECT       tr_id
  ,       start_date, end_date          -- If wanted
  ,       24 * ( ( ( TRUNC (end_date,   'IW')          -- Count -1 day for every full week
                      - TRUNC (start_date, 'IW')
               )
             / -7
                )
              + LEAST ( end_date               -- If end_date is a Sunday
                          , TRUNC (end_date, 'IW') + 6     -- consider it 00:00:00 on Sunday
                   )
              - CASE
                        WHEN  start_date >= TRUNC (start_date, 'IW') + 6     -- If start_date is a Sunday
                 THEN  TRUNC (start_date, 'IW') + 6               -- consider it 00:00:00 Sunday
                 ELSE  start_date
                    END
              )     AS total_hours
  FROM      got_extrema
  ;
  

  I guess that you don't need to worry about fractions of a second. As you did in the code you have posted, I am to convert the TIMESTAMP to date values, because of DATE arithmetic and functions are so much better than what is available for timestamps.

  Basically, it's just to find the number of days between start_date and end_date and multiplying by 24, with these twists:
  (a) 1 day is deducted for each week between start_date and end_date
  (b) if End_date is a Sunday, none of the end_date himself hours are counted
  (c) If start_date is a Sunday, then all the start_date himself hours are counted. Why these hours should be counted? Because 1 day is already being deducted for the week which includes start_date, which contains only this Sunday.

  TRUNC (dt, 'IW') is the beginning of the ISO week containing dt; in other words, 00:00:00 the or before the dt last Monday. This is not the NLS parameters.

  Of course, I can't test without some sample data and the exact results you want from these data. You may need a little something more If start_date and end_date are both on the same Sunday.
  Whenever you have a problem, please post a small example of data (CREATE TABLE and only relevant columns, INSERT statements) of all of the tables involved.
  Also post the results you want from this data, as well as an explanation of how you get these results from these data, with specific examples.
  Always tell what version of Oracle you are using.
  See the FAQ forum {message identifier: = 9360002}

• How to find the difference between two dates in the presentation layer

  Hi gurus,
  
  Hello to everyone. Today, I came with the new requirement.
  
  
  I need to know the difference between a date and the current date in the formula column application presentation layer.by.
  
  
  
  Thank you and best regards,
  Prates

  Hi Navin,

  TIMESTAMPDIFF function first determines the timestamp component that corresponds to the specified interval setting. For example, SQL_TSI_DAY corresponds to the day component and SQL_TSI_MONTH corresponds to the component "month".

  If you want to display the difference between two dates in days using SQL_TSI_DAY, unlike butterflies SQL_TSI_MONTH and so on...

  hope you understand...

  Award points and to close the debate, if your question is answered.

  See you soon,.
  Aravind

• How to find the difference in date in years, months and days

  Hello
  I need to find the difference between 2 dates in the following way "years months days". "
  Please can help me, how can I achieve this.
  
  for example, in the scott schema emp table, I need to find the difference in date between sysdate and hiredate for an employee in the following way.
  
  12 years 7 months 4 days.

  Hello

  Please, see this post to AskTom [difference between 2 dates | http://asktom.oracle.com/pls/asktom/f?p=100:11:0:P11_QUESTION_ID:96012348060]. There is good information in this forum, for example you can also see this thread [calculation of years, months & days | http://forums.oracle.com/forums/thread.jspa?messageID=3115216�]

  Kind regards

  Published by: Walter Fernández on November 30, 2008 08:58 - adding another link