Added two Dates and number of days

Similar Questions

• Find the difference between two date and time

  Hi friends,
  
  I wanted to find the difference between two date and time, but my qury is slightest error "invalid number."
  
  

  select sql_step_num,proc_name,run_seqno,start_date,end_date,(to_char(start_date,'HH24-MI-SS') - to_char(end_date,'HH24-MI-SS') ) as ed  
  from eval.EVAL_RUNTIME_DETAILS
  where trunc(start_date) = trunc(sysdate) 
  order by sql_step_num;

  You try to get the feel between two char strings.
  And more difference between two dates gives a NUMBER of days.
  Try this:

  select sql_step_num,proc_name,run_seqno,start_date,end_date,numtodsinterval(end_date-start_date,'DAY') as ed
  from eval.EVAL_RUNTIME_DETAILS
  where trunc(start_date) = trunc(sysdate)
  order by sql_step_num;
  

• How to compare two dates and find exactly

  How to compare two dates and find the exact age of the person, no one could be an age of child 2 days or a month, or other.

  I'd really appreciate if someone help o

  Concerning

  After spending 2 hours, I go out with a solution by myself, how ever the function can be customize to check if the user enters date right.

  function findAge(subjectName,fromdate, todate) {
      console.log("findAge(fromdate, todate) is called now "+subjectName+"-->"+fromdate+"-->"+todate);
      if(todate) todate= new Date(todate);
      else todate= new Date();
  
      var age= [], fromdate= new Date(fromdate),
      y= [todate.getFullYear(), fromdate.getFullYear()],
      ydiff= y[0]-y[1],
      m= [todate.getMonth(), fromdate.getMonth()],
      mdiff= m[0]-m[1],
      d= [todate.getDate(), fromdate.getDate()],
      ddiff= d[0]-d[1];
  
      if(mdiff < 0 || (mdiff=== 0 && ddiff<0))--ydiff;
      if(mdiff<0) mdiff+= 11;
      if(ddiff<0){
          fromdate.setMonth(m[1]+1, 0);
          ddiff= fromdate.getDate()-d[1]+d[0];
          --mdiff;
      }
      if(ydiff> 0) age.push(ydiff+ ' year'+(ydiff> 1? 's ':' '));
      if(mdiff> 0) age.push(mdiff+ ' month'+(mdiff> 1? 's':''));
      if(ddiff> 0) age.push(ddiff+ ' day'+(ddiff> 1? 's':''));
      if(age.length>1) age.splice(age.length-1,0,' and ');
      console.log("===============================");
      console.log("Subject age is = "+age.join(''));
      console.log(" age Day = "+ddiff);
      console.log(" age Month = "+mdiff);
      console.log(" age Year = "+ydiff);
      console.log("===============================");
      var subjectAGE =  age.join('');
  
  }
  

  peardox Thanks for the reply

• How to compare two dates and times in BPEL?

  Hi all
  I need to compare two dates and times in a switch activity, but could not find any function for her.
  
  My switch like this activity:
  
  case Date1 > date2:
  do something;
  otherwise:
  do something;
  
  He is not such a function in 'The functions of Date' and can not find this function to xpath.
  
  IBE, there is no function for a period of time to make (less one date of another?...) If this return to the data compare question).
  
  Thank you.

  Hello

  Logic function as more/less function can be used to compare two dates so that same comapring two numbers.

  Ex: If your variable reception contains two pieces of data such as date1 and date2, then you can use format in condition switch below.

  XpathXpression (date1) > XpathXpression (date2)

  hope this will help you.

• 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.

• I need to get date-date by number of days, hours, minutes, seconds.

  I have two dates, A and b... .i need to get the B - number of days, hours, minutes, seconds.
  
  I need differnce between two dates in days, hours, minutes, seconds.
  
  the day was 8 hours. Sat/Sun off
  
  It's simple if we consider our regular calendar... .but how can I get the same if a work_calander...
  
  That is, for example, A = Fri evening 17:00
  B = Monday evening 17:00
  
  the result should be 8 hours... (Sat/Sun, judge 9-6 to the timetables of office.. .so 1 pm Friday and 7: 00 in the LUN)
  
  such a schedule is defined.
  
  Please give a solution
  
  Thanks in advance
  Prince
  
  Published by: Prince on August 27, 2008 23:42

  your query is not clear enough.

• Current Date more number of days script

  Hi all

  I have a form that I'm building that has a date field (currently Setup to specify the date of today on the load).  There is a field that slects number of days until a billing cycle (currently 0 in 13 days).

  I need to set up a script that takes today's date, adds the number of days since the result of 0-13 and gives a result of a date (0-13 days from today for example).

  Any thoughts on how I could do this?  Very new to JS.

  Thanks for any help you may be able to provide.

  Ant

  Here is an example of what I used:

  var f = this.getField ("BILLINGCYCLEDATE");

  G1 var = this.getField ("DAYS");

  var g = g1.value

  var today = new Date();

  Add var = d1.valueOf ();

  Add += 1000 * 60 * 60 * 24 g,

  future var = new Date (add);

  f.Value = util.printd ("mm/dd/yyyy", future);

  You do not want to set up as a calculation in your field BILLINGCYCLEDATE.  Field DAYS, that's all that you configured as the counter 0-13.  I implemented a simple text box where you enter a number.  The field will then calculate X number of days.   I hope this helps.

• Adding a date and time to create a channel in a data plugin

  I would like to preface this question with the fact that I am new to VBS programming.  What I want to do is to create a plugin. Most of it is done, but I'm having a problem with the addition of two times to a.  I have a text file that stores the date and time separated by a tab character.  This file format was designed to excel.  When I bring him both the date and time are in different channels and I can not figure out a way to add both and either replace them or even place it in a separate variable.  Here is an example.

  Chan (1) contains the date, Chan (2) contains both.  This statement, which places the data in the string is:

  The StandChannels value (0). Values (k) = Chan (1). Values (i)

  I tried:

  The StandChannels value (0). Values (k) = Chan (1). Values (i) + Chan (2). Values (i)

  error: object doesn't support this property or method

  or

  Chan (1). Values (i) = Chan (1). Values (i) + Chan (2). Values (i)

  error: object doesn't support this property or method

  I tried several other iterations and have had no success.  This seems to be an easy task, Miss me probably something simple.

  Hi Eric,.

  The first problem is that your datetime format string was wrong, you need to use:

  File.Formatter.TimeFormat = "MM/DD/YY SS.fff.

  Or maybe if your year is first of all it must be:

  File.Formatter.TimeFormat = "YY/MM/DD SS.fff.

  The second question was slippery.  The newsgroups (0) you throw combines the ProcessedChannel values correctly, but you were not successfully picking off those that you wanted to leave Chan (0).  I don't really know why the method you were using did not work, but I found a similar one that does.  Note that I've only changes part SUB New_File (file, ChanNames) of the code, not the SUB Previous_File (file, ChanNames) part of the code (got tired of typing).

  Brad Turpin

  Tiara Product Support Engineer

  National Instruments

• Subtracting between two dates and see the result as a date

  HII everyone,
  
  I had to create a procedure that checks the difference of dates in one month...
  I've got is the view of the difference in the form of date itself...
  
  
  for example:
  
  01/03/2012
  01/03/2012
  01/03/2012
  01/05/2012
  01/05/2012
  .
  .
  so now for a full month...
  
  the output should be as data are missing for the date 01/04/2012...
  is it possible to display the output as given...
  I am new to oracle, help would be much appreciated...
  Thank you..
  
  Published by: 948895 on August 2, 2012 21:39

  To fill the gaps between the two dates:

  select &&start_date + (level - 1)
  from dual
  connect by level <= (&&end_date - &&start_date)
  /
  

  To generate the dates for one month:

  select trunc(&&start_date, 'MM')) + (level - 1)
  from dual
  connect by level <= to_number(to_char( last(day(&&start_date)))
  /
  

  This example uses SQL * more variable substitution, but you can easily change to use the value of PL/SQL or everything that you need.

  Cheers, APC

• help needed by subtracting a line that contains (date and time) + business days

  Hi all
  Could someone please help me reg this
  I have a single column including (date and time) for example, 28/03/2009-12:40
  I want to calculate the days of work for an employee
  the table is as follows
  Opn_in_out OPN_MOV_DATE name
  
  2009-03-28 12:40 abc 2
  2009-03-29 08:25 1 abc
  2009-03-28 16:15 2 def
  2009-03-29 08:45 1 def
  
  '2' means is out for work
  '1' means employee came after work
  ABC and def are the different employees, how to calculate the days each has worked, I tried date_diff but it does not work
  Please help me
  concerning
  userconfused

  As explained above...
  you need to replace...

  (select case when tot>3 then 1 end as days ) 
  

  with a little (even in brackets are not required)

  case when tot>3 then 1 end as days 
  

  Ravi Kumar

• Rename the files in a directory by adding the date and time

  Hello

  I am looking for a way to rename all the files that are in a directory with an add-on at the end with the date and time.

  for example:

  File1-> file1ddmmyyhhmi

  File2-> file1ddmmyyhhmi

  for example:

  $A = get-date-Format "yyyymmddhhmm.

  foreach-object - process ...

  Thank you.

  You can try something like this

  $A=get-date -Format "yyyymmddhhmm"
  Get-ChildItem "C:\folder" | %{
       Rename-Item -NewName ($_.Basename + $A + $_.extension) -Path $_.FullName
  }
  

  This assumes that the files are stored in C:\folder

  ____________

  Blog: LucD notes

  Twitter: lucd22

• NG Date and number fields internationalization

  We develop forms PDF for a format of number of spectators in Switzerland (Swiss) German is 99 999,99 ' and JJ.mm.aaaa.

  How do we set the format required for number fields and date?

  That's why there is the option 'Custom' for the 'number' and 'Date' format of the options. You may need to convert the field values for the numbers to use the "." decimal point.

• EXTRACT DATES WITH THE HELP OF TWO DATES AND THE BOX

  Hello
  
  I HAVE 2 DATE PICKER ITEMS
  : P1_ARRIVAL_DATE AND: P1_DEPARTURE_DATE
  
  AND I HAVE 7 ITEMS in BOXES DAY it return values are 'Y '.
  
  : P1_MON
  : P1_TUE
  : P1_WED
  : P1_THU
  : P1_FRI
  : P1_SAT
  : P1_SUN
  
  as I choose date: P1_ARRIVAL_DATE AND: P1_DEPARTURE_DATE
  AND THE DAY, AS: P1_MON, ONLY THE DATES HAVE TO RETURN WHO HAVE MONDAY AS DAY BETWEEN: P1_ARRIVAL_DATE AND: P1_DEPARTURE_DATE.
  
  
  FOR EXAMPLE IF I CHOOSE THE DATE: P1_ARRIVAL_DATE AS 07_NOV_2011 AND: P1_DEPARTUE_DATE AS: P22_NOV_2011 AND IT HAS 3 MONDAY BETWEEN THESE DATES
  RESULT SHOULD BE DISPLAY AS
  ======================================

   ARRIVAL DATE-------ARRIVAL DATE+6
  ============================
   07_NOV_2011 ---     13_NOV_2011       // THESE ARE MONDAY DATES
  14_NOV_2011 ---     20_NOV_2011     
  21_NOV_2011 --      28_NOV_2011 

  CAN I CHOOSE MORE THAN ONE CHECK BOX AT THE SAME TIME
  FOR EXAMPLE IF I CHOOSE: P1_MON AND: P1_TUE, THE RESULT SHOULD BE
  
  

  ARRIVAL DATE-------ARRIVAL DATE+6
  ============================
   07_NOV_2011 ---     13_NOV_2011   
   08_NOV_2011 ---     14_NOV_2011      
  14_NOV_2011 ---     20_NOV_2011      // MONDAY AND TUESDAY DATES
  15_NOV_2011 ---     21_NOV_2011  
  21_NOV_2011 --      27_NOV_2011
  22_NOV_2011 ---     28_NOV_2011 

  HOW CAN I DO THIS? PLEASE HELP SOLVE THE PROBLEM.
  
  
  
  THANKS and GREETINGS
  CORINE

  X the problem you see is a misunderstanding on my part about what is the date of arrival date of departure... it's inverse reasoning: at your arrival is before departure (and I got it on the other hand, the departure before arrival (:-))

  But there is also a flaw in the code (instead of the Date of arrival by using the select sysdate).

  Please run the code like this:

  with dates as
  (select to_date(:P1_ARRIVAL_DATE,'DD-MON-YYYY')+level-1 as day from dual
   connect by level <= to_date(:P1_DEPARTURE_DATE,'DD-MON-YYYY') - to_date(:P1_ARRIVAL_DATE,'DD-MON-YYYY')  +1
   )
  select day, to_char(day,'Dy') from dates
  where to_char(day,'Dy') = 'Mon' and :P1_MON = 'Y'
  union
  select day, to_char(day,'Dy') from dates
  where to_char(day,'Dy') = 'Tue' and :P1_TUE = 'Y'
  union
  select day, to_char(day,'Dy') from dates
  where to_char(day,'Dy') = 'Wed' and :P1_WED = 'Y'
  union
  select day, to_char(day,'Dy') from dates
  where to_char(day,'Dy') = 'Thu' and :P1_THU = 'Y'
  union
  select day, to_char(day,'Dy') from dates
  where to_char(day,'Dy') = 'Fri' and :P1_FRI = 'Y'
  union
  select day, to_char(day,'Dy') from dates
  where to_char(day,'Dy') = 'Sat' and :P1_SAT = 'Y'
  union
  select day, to_char(day,'Dy') from dates
  where to_char(day,'Dy') = 'Sun' and :P1_SUN = 'Y'
  

  for the period from 01 to 15 November 2011, with P1_TUE and P1_FRI, flag in there, I got this result:

  DAY      TO_CHAR(DAY,'DY')
  11/01/2011     Tue
  11/04/2011     Fri
  11/08/2011     Tue
  11/11/2011     Fri
  11/15/2011     Tue
  

  To simplify (can only test inside the apex report, could not run code in SQL Wkshop):

  with dates as
  (select to_date(:P1_ARRIVAL_DATE,'DD-MON-YYYY')+level-1 as day from dual
   connect by level <= to_date(:P1_DEPARTURE_DATE,'DD-MON-YYYY') - to_date(:P1_ARRIVAL_DATE,'DD-MON-YYYY') +1
   )
  select day, to_char(day,'Dy') from dates
  where v('P1_'||to_char(day,'DY')) = 'Y'
  

  Published by: Kléber M on November 14, 2011 05:06

• 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");
  //

• You will need to add the number of days to the end date of the users.

  Hello
  I have a code where we adding the number of days (30) to the current date and update IDM DB user end date.
  Now, we have a requirement to add the number of days (30) to the existing fence of user instead of add to the current date.
  
  * public String incrementDate (int daysToAdd)
  {
  Start date
  log.info("NotifyLastDayOfService::incrementDate(): entrez»);
  SimpleDateFormat sdf = new SimpleDateFormat ("yyyy-MM-DD 00:00:00");
  Calendar c = Calendar.GetInstance ();
  c.Add (Calendar.DATE, daysToAdd); number of days to add
  String PROMOTIONDate = sdf.format (c.getTime ());
  log.info("NotifyLastDayOfService::incrementDate(): sortie»);
  return Nouvelle_date;
  }*/
  
  Have any body set up this scenario?
  Please suggest.
  
  
  Thank you
  Kalpana.

  You can try this as well

  where oimAttribute = "Users.End Date"

  public String updateOIMAttributeForDateType (String oimAttribute, Date endDate, long take, String daysToAdd)
  {
  tcUserOperationsIntf userOperationsIntf = null;
  HashMap modifyMap = new HashMap ();
  HashMap searchmap = new HashMap ();
  SimpleDateFormat formatter = new SimpleDateFormat ("yyyy-MM-DD hh: mm :"); ")
            
  try {}
  Calander cal = null;
  cal.setTime (endDate);
  Cal.Add (Calendar.DATE, Integer.parseInt (daysToAdd));
  Date Edate = cal.getTime ();

  modifyMap.put (oimAttribute, Edate);
  searchmap.put ("Users.Key", String.valueOf (take));
  userOperationsIntf = Platform.getService (tcUserOperationsIntf.class);
  tcResultSet searchset = (searchmap) userOperationsIntf.findAllUsers;
  userOperationsIntf.updateUser (searchset, modifyMap);
  return a SUCCESS;
  } catch (Exception e) {}
  If (Logger.isErrorEnabled ()) logger.error (e.getMessage ());
  error return;
  } {Finally
  try {}
  If (userOperationsIntf! = null) userOperationsIntf.close ();
  } catch (Exception e) {if (logger.isErrorEnabled ()) logger.error (e.getMessage ()) ;}}
  }