Leap year Date issue (example)

Similar Questions

• Wrong date. A day off. Does not take into account leap years.

  We don't want the date on my iMac. It is a day of rest. On March 3, it shows March 2. Of course, it didn't take leap years into account. When I tried to change it after you have unlocked the date/time, he wouldn't let me to change the date. It is automatically defined under Apple Americas/U.S. (time.apple.com site), but the automatic setting is false. How do override us the auto setting or have them repaired?

  Oops! My watch was the one who was wrong. Apple is very well. Sorry about that.

• Leap year TMS Scheduler

  We have MSD and MSD Smart Scheduler.  When users of planning meetings in the smart Scheduler for Monday 2/29, they are listed as Tuesday 3/1, I guess because of the leap year.  However, it shows the correct date when you click the meeting and in the TMS itself.  I checked to make sure that the time zone settings are correct.  Is that going to the issue of Monday?

  Turns out just a cosmetic defect with Smart Scheduler, see CSCuy19844.

• How to check leap year between the years earn a new degree?

  Hi all
  I developed a report to print a report of emplyees who earn a new degree between towing dates entered by two parameters of the user.
  I used this code to do this:
  
  SELECT ALL E.EMP_SID, E.EMP_FULL_NAME, E.FINANCIAL_NO
  (CASE WHEN E.qualification_sid = 1104 AND E.degree_sid = 8 THEN
  (TO_DATE (E.degree_DATE, 'DD-MM-RR') +(365*M.spe_per) + 1).
  ON THE OTHER
  (TO_DATE (E.degree_DATE, 'DD-MM-RR') +(365*M.yea_per) + 1).
  END)
  COMM_degree,
  OF INDUSTSRIE E, M DEGREES
  WHERE (CASE WHEN E.qualification_sid = 1104 AND E.degree_sid = 8 THEN
  (TO_DATE (E.degree_DATE, 'DD-MM-RR') +(365*M.spe_per) + 1).
  ON THE OTHER
  (TO_DATE (E.degree_DATE, 'DD-MM-RR') +(365*M.yea_per) + 1).
  END)
  
  BETWEEN TO_DATE (: P_F_DATE, 'DD-MM-RR') AND TO_DATE (: P_L_DATE, 'DD-MM-RR')
  AND (E.degree_SID = M.SID)
  ORDER BY M.SID DESC, E.EMP_FULL_NAME
  
  Now there is an error in the data led to some employees with a leap year in the year to earn income
  for example: an employee has a degree date of 10/01/2011 and win it supposed the next level is 10/01/2016 but the report gives as 30/09/2016 because there's one year leap year 2012 which have 366 days 365 days I count in the code!
  If there is something mysterious in the code, please do not hesitate to ask!
  Please how to fix this?
  
  Note: I use reports 6i.

  Hello

  Try ADD_MONTHS(SYSDATE,12) - instead of 365 DAYS

  IF ANY PROBLEM WE THE KNOW

• interval + "1" [year | month | day] does not take into account leap years?

  Hi all
  
  The Version of database: Oracle Database 10g Enterprise Edition Release 10.2.0.5.0 - 64bi
  Database version: PL/SQL Release 10.2.0.5.0 - Production
  
  I just realized that the function of the interval is not account of leap years.
  

  SQL> select to_date('29-02-2008','dd-mm-yyyy') + interval '1' year 
    2    from dual;
  select to_date('29-02-2008','dd-mm-yyyy') + interval '1' year
                                        *
  ERROR at line 1:
  ORA-01839: date not valid for month specified

  I looked through the documentation as much as I can and could find no mention of this.
  
  If I want to add a year to date, I have to calculate the number of days or do something like incrementing only the part of the year and then manually check that it does not fall on February 29 if it is not a leap year manually?
  
  Edit:
  
  just to clarify, day and minute seem to work for example but do not have month and year:
  

  SQL> select to_date('28-02-2009','dd-mm-yyyy') + interval '1' day
    2    from dual;
  
  TO_DATE('
  ---------
  01/MAR/09
  
  1 row selected.
  
  Elapsed: 00:00:00.00
  SQL> 
  SQL> select to_date('28-02-2009 23:59','dd-mm-yyyy hh24:mi') + interval '1' minute 
    2    from dual;
  
  TO_DATE('
  ---------
  01/MAR/09
  
  1 row selected.
  
  Elapsed: 00:00:00.00
  SQL> 
  SQL> select to_date('29-01-2009','dd-mm-yyyy') + interval '1' month
    2    from dual;
  select to_date('29-01-2009','dd-mm-yyyy') + interval '1' month
                                        *
  ERROR at line 1:
  ORA-01839: date not valid for month specified
  
  
  Elapsed: 00:00:00.00
  SQL> 
  SQL> select to_date('29-02-2008','dd-mm-yyyy') + interval '1' year
    2    from dual; 
  select to_date('29-02-2008','dd-mm-yyyy') + interval '1' year
                                        *
  ERROR at line 1:
  ORA-01839: date not valid for month specified

  Published by: WhiteHat on July 14, 2011 10:31

  Hello

  It is true; in the INTERVAL arithmetic + x (or) d - x, where x is the months or years, the result will be the same day of the month as d. If this date does not exist, then you will get an error
  Another example is:

  SQL> SELECT  DATE '2011-03-31' + INTERVAL '1' MONTH  FROM dual;
  SELECT  DATE '2011-03-31' + INTERVAL '1' MONTH  FROM dual
                            *
  ERROR at line 1:
  ORA-01839: date not valid for month specified
  

  To avoid triggering an error, use ADD_MONTHS and/or arithmetic DATE instead.

• Bug or error during the handling of leap years with WDS/DS

  Hi all
  
  I discovered a difference in results with the construction and to debug (without frame) and generate and debug with the revisions of Web or server of determinations when calculating the days in leap years.
  I use the DayDifferenceInclusive function to calculate the number of days in a period of insurance.
  
  the number of days of insurance period = DayDifferenceInclusive (the start date of the insurance, the date of the end of insurance)
  
  With the tester of regression or generation and debugging (without frame), I get the following result for the following entry:
  
  * 93 = DayDifferenceInclusive (02-29-2008, 05-31-2008) *.
  
  When we use the revisions of Web or server of determinations we get the value of * 92 * for exactly the same data entry when it should be 93.
  Is this a bug in the java code? How can we fix this or is there a work around it?
  
  We are in the middle of integration testing system for a huge project to help is very appreciated!
  
  Best regards
  Niels Roest

  Niels,

  Thanks for that - and it seems that in the Netherlands DST began March 30, 2008, that is, as far as I know, the trigger for this error.

  I have reproduced the issue against 10.1 using the period from 1st October 2010 to 30 October 2010, which contains the date this summer time began in Australia.

  Regarding the ways to work around the problem, I can offer 3:

  * wait 10.2 - it is in beta at the moment (but given your opening remarks, may not be an option for you)
  * set your locale of the server to that which has no DST
  * work around the issue in the rules:

  First of all, the question is that the retail version fails because it does not realize that one of these days is shorter than the standard 24 hours.

  The trick is to replace the DayDifference function with the HourDifference function and round up the number of days.

  Thus given a rule that looks like:

  result = DayDifference (date one, date two)

  It can be rewritten in the form:

  The result = round (HourDifference (one at 00:00:00, date two to 00:00:00) / 24, 0)

  * the "at 00:00:00 ' is necessary because HourDifference works on dates/hours, not dates.

  Sorry for the inconvenience.

  Concerning
  Andrew

• Year (date) and Month (date) "YYMM" in number.

  

  In the column 'G' I use the formula = month (date)

  In the column 'H' I use the formula = year (date)

  Is it possible to use the formula above to create a "YYMM' as the column 'I '?

  Column 'G and H' will be deleted if the column 'I' is the work.

  Hi LV.

  Here is an example of use of Numbers ' 09. The first method (in column (C) is feasible in 3 numbers; the second is maybe not.

  

  Columns C and D are left in the default auto format (and alignment).

  Column C, automatically align left, excerpts from the last two formatted numbers Year() then extracts and oncatenates the last two digits of a string built consisting of two zeros & the number Month() one or two. The result is a text string of four digits, as shown.

  C2 and filled down: = RIGHT ((B), 2 YEAR) & RIGHT ("00" & MONTH (B), 2)

  Column D simply copies the string Date and time of B and displays it by using a custom format, showing that the two-digit year & the monthnumber double-digit.

  D2 and filled down: = B

  Custom format for the cells D2 - D8:

  

  Kind regards

  Barry

• Zero-view fixed year date do?

  Zero-view fixed year date do?

  Refer to this link, which explain the zero display feature for example.

  Hyperion Financial Management: Zero display and default display settings. Edgewater Ranzal Weblog

  Let us know if you more details.

  Thank you

  ~ KKT ~.

• Bug with interval and leap years

  select to_date('2012-feb-29','yyyy-mon-dd') + interval '1' year as dt from dual;
  
  ORA-01839: date not valid for month specified
  01839. 00000 -  "date not valid for month specified"
  *Cause:    
  *Action:
  
  select to_date('2012-feb-29','yyyy-mon-dd') + interval '2' year as dt from dual;
  
  ORA-01839: date not valid for month specified
  01839. 00000 -  "date not valid for month specified"
  *Cause:    
  *Action:
  
  select to_date('2012-feb-29','yyyy-mon-dd') + interval '3' year as dt from dual;
  
  ORA-01839: date not valid for month specified
  01839. 00000 -  "date not valid for month specified"
  *Cause:    
  *Action:
  
  select to_date('2012-feb-29','yyyy-mon-dd') + interval '4' year as dt from dual;
  
  29-FEB-16 00:00:00
  
  
  select to_date('2012-feb-29','yyyy-mon-dd') + interval '1' day as dt from dual;
  
  01-MAR-12 00:00:00
  
  select to_date('2012-feb-29','yyyy-mon-dd') + interval '1' month as dt from dual;
  
  29-MAR-12 00:00:00

  The problem exists in 10.2.0.4 and 11.2.0.3
  
  Published by: birlenbach-Chase February 29, 2012 09:20

  Hello

  That's just how the intervals of work. Leap years are the least of the problem; Adding 1 month to 31 March results in the same error.
  If you want to make sure that the result is a valid DATE, use ADD_MONTHS. (There is no separate function for adding years; use ADD_MONTH (SYSDATE, 12 * n) to get the DATE which is n years.)

• Is there a tag to display the current year? For example, the exercise of copyright?

  Is there a tag to display the current year? For example, the exercise of copyright? I want to put this in the site map (I guess) so it is displayed on the footer of each page.

  With the addition of liquid you can easily do this using:

  {{globals.site.dateNow | date: "yyyy"}}

  Please review the documentation to learn more about what you can achieve with liquid:

  Introduction to liquid

  Liquid | Reference

• How handle the date of the leap year in the query?

  Dear Exparts
  Hope you will be fine.
  Here is my banner

  Oracle Database 10g Enterprise Edition Release 10.2.0.3.0 - Prod

  I have a date field name birth_date where I stored the date of birth. I need to choose the date of birth in a given range. For example

  BIRTH_DATE
  -------------------------
  01-FEB-1980
  12-FEB-1984
  29-FEB-1992
  01-MAR-1986
  01-APR-1986
  22-APR-1988
  01-MAR-1992

  The setting of my research is

  brith_date between 01/02/13 and 01/04/13

  I want the file

  01-FEB-1980
  12-FEB-1984
  29-FEB-1992
  01-MAR-1986
  01-APR-1986

  Here is the script
  create table

  CREATE TABLE DATE_OF_BIRTH 
  (
    BIRTH_DATE DATE 
  );

  inserting data

  insert into DATE_OF_BIRTH 
  values(to_date('01-FEB-1980','DD-MON-RRRR'));
  insert into DATE_OF_BIRTH 
  values(to_date('12-FEB-1984','DD-MON-RRRR'));
  insert into DATE_OF_BIRTH 
  values(to_date('29-FEB-1992','DD-MON-RRRR'));
  insert into DATE_OF_BIRTH 
  values(to_date('01-MAR-1986','DD-MON-RRRR'));
  insert into DATE_OF_BIRTH 
  values(to_date('01-APR-1986','DD-MON-RRRR'));

  Thanks in advance... (Thanks to Frank Kulash)
  
  ask2Learn
  
  Published by: asked to learn, on 24 February 2013 19:40

  select  *
    from  date_of_birth
    where to_char(birth_date,'mmdd') between '0201' and '0401'
  /
  
  BIRTH_DAT
  ---------
  01-FEB-80
  12-FEB-84
  29-FEB-92
  01-MAR-86
  01-APR-86
  
  SQL>
  

  SY.

• Question of TIMESTAMPADD (SQL_TSA_YEAR) (with leap years)

  Dear community OBIEE,

  Any who have suffered no problems when you use timestampadd (SQL_TSI_YEAR, 55, 'Date of birth')?  Most of the result were correct with the exception of a few minute question when a DOB fell 2/28/20... (number pair year). Is there a work around for a such minute but difficult behavior OBIEE?

  Kind regards.

  -Felix

  For me, it's showing as expected.

  See below. in my case I have not value date 28/02/1977 so formulated according to the needs. Maybe this would be a reason to see the 28/02/2032

  

• How can I tell Lightroom 6 I want to use years/dates as my file folder names

  Recently I bought Lightroom, however I have thousands of images in Photoshop Elements.   I did not use years and dates that the principal names for folders for organizing.  You want to continue the system that I so need to change the defaults in Lightroom 6.  Looked for centuries and can not find where to change it.  Also have tried on this forum and Google search, but no significant results.  How to do this please?

  It is not different in LR 6 / CC in LR 5.  You can find it in the dialogue right hand panel.  You may need to click the small arrow just to the right of the word 'Destination' to see.

• How to compare a Date with the current year data and member of period on FIX

  Hi experts,
  
  I have a Sun project that each Member is a project (P01, P02,...)
  and account IMGs who stores the information of each project (name, Start date, finished...)
  Finished Date member is of data type Date
  
  So how can write an IF condition below in order to be able to compare the end Date of the project with the members of the current year and period on FIX
  DIFFICULTY (@Descendants (Projects), Descendants (year-round), Descendants ("year Total")...)
  IF (@CURRMBR (Period)-> @CURRMBR (Year) < Project > - FinishedDate)
  Do something...
  On the other
  Do something
  
  
  Please help me on this. Sorry for my bad grammar. Please ask if there is something not clear
  
  Thank you very much
  Van Huy.
  
  Published by: Van Huy on 29 January 2013 01:14
  
  Published by: Van Huy on 29 January 2013 02:24
  
  Published by: Van Huy on 29 January 2013 02:25
  
  Published by: Van Huy on 29 January 2013 18:04

  Here's what I did. Post for which may later cover

  VAR FM; / * End of month of the project.
  VAR EXERCISE; / * Year of the project terminated * /.
  VAR CM; / * Capture the current month on FIX declarations * /.
  VAR CY; / * Capture current year on FIX declarations * /.

  DIFFICULTY (@RELATIVE ('year', 0), @RELATIVE ('period', 0), @IDescendants ("core projects")...)

  FY = @ROUND ('TGHT'-> ' NA contract "-> 'FY06'-> ' NA period" / 10000, 0);
  FM = @MOD (@ROUND ('TGHT'-> ' NA contract "-> 'FY06'-> ' NA period" / 100, 0), 100);
  / * For FY13 back 13... */
  CY = @JgetDoubleFromString (@CONCATENATE ('20', @SUBSTRING (@NAME (@CURRMBRRANGE (year, Lev, 0, 0, 0)), 2)));
  / * Value CM set based on perceived period on the DIFFICULTY of statement * /.
  IF (@ISMBR ("Jan"))
  CM = 1;
  ELSE IF (@ISMBR ("February"))
  CM = 2;
  .....
  ELSE IF (@ISMBR ("Dec"))
  CM = 12;
  ENDIF

  IF (CY< fy="" or="" (="" cy="=" fy="" and="" cm=""><>
  Do something...
  ON THE OTHER
  Do something...
  ENDIF
  ENDFIX

  Published by: Van Huy on 19 February 2013 23:10

  Published by: Van Huy on February 20, 2013 19:46

• Import-CSV and Date issues

  OK, I have a CSV file that I store a lot of information about the virtual machines as the name of the applicant, decommission date, server name, etc., etc.

  I am trying to create a script to automate the process of decommissioning of auto generate an e-mail to the applicant with the name date and decom server to see if they want to extend the lease of the server or delete. I'm having some trouble with the cmdlt of Get-Date however. I would like to run this script once per week via windows scheduled task. I want to target VM decomission whose date is within 2 weeks.

  Is there something I'm missing here.

  $Date = (get-Date - DisplayHint Date). AddDays (14) every time I do that it adds the time...

  $decom_info = import-CSV Book.csv

  {foreach ($project to $decom_info)

  if($Project.) Decom_Date - ilt $Date) {}

  Send email code

  }

  For some reason any this is incorrect however. I expect because the "Decom_Date" column is pulled into a string, or something. Any ideas?

  The date the field of your reading of the CSV file is a string.

  But you easily "cast" it is in an object DateTime.

  See the following example.

  $t = "09/21/2010"
  $t.GetType()
  $s = [datetime]$t
  $s.GetType()
  

  Because there are some hooks in the code and the SW forum don't like these, I have attached the code

  ____________

  Blog: LucD notes

  Twitter: lucd22