month days years

Similar Questions

• Dates on Firefox don't appear in the American Style, that is, month day year I want to change to day month year please help?

  Could you please tell me how to change the display of the date to display the day month year instead of month day year?

  HOW to: Change the Date, time, number, and currency value displays in Windows XP

  http://support.Microsoft.com/?kbid=307938

  Check and tell if its working.

• Date format Conversion MM/DD/YYYY month day, year

  Hi all
  
  Is there a BEEP function that converts the DD/MM/YYYY month day, year?
  
  for example.
  
  08/10/2010 convert to 8 October 2010
  
  Thank you very much!

  Since your date is not in canonical format, you cannot format dates directly.
  You will have to do with workarounds.
  Try to use
  ,

  Here, I assume that your date is delivered in the format MM/DD/YYYY

• What is the date format used in the charts of Microsoft Support policy? Is the day/month/year or month/day/year?

  For example, the Date of beginning of life cycle shows 31/05/2001 (March 5) and the end of Support Date extended watch 07/12/2001 (12 July or December 7).

  The date format is month/date/year. This format is for the start date and end date for extended Support.
  In your example, the start date is May 31, 2001, and the end date would be July 12, 2001.

  I hope this helps.

  Marilyn

• Can I change the date format day/United States (month day year) in the United Kingdom (date month year)?

  When I consult databases via Firefox, the format of date/day is set at: day, month, year (AMERICAN style), while I need, day, month, year, I work in the United Kingdom and the 'bad' format is VERY confusing and could lead to errors and mistakes. IE has that format, but I don't want to return to using that.

  You use the English (United States) version of Firefox ESR 24.3.0. You can download the English version of (British) below:

  • http://www.Mozilla.org/Firefox/organizations/all/

  1. under Options - content - language, make sure that English/United Kingdom [en-us] is located at the top of the list.

  • Web content settings, pop ups, fonts and languages

  2. in the Windows Control Panel category, region and language, make sure that the short Date is set at your convenience.

  • http://Windows.Microsoft.com/en-us/Windows/change-dates-times-currency-measurements-display

  You can check the date and time at the library of Firefox, bookmarks category format. Press Ctrl + Shift + B to open it, and then note the dates in the added to the right column.

• I want to collect rain for the day, month and year; Can what formula I use?

  I want to collect rain for the day, month and year; What formula or expression that I can use.

  I use a Rain Wise product that converts impulses to an analog value.  The rain wise device can be

  "the value measure up to 1", 5", or 10".  I will be setting the unit at 10 inches in increments of 0.01 inch.

  What I would do is at each time increments of signal I would consider it as 0.01 then after reaching

  a period of time a day or a month, back to zero.

  Need some advice on this problem.

  How about you, using a counter to count the number of times where the signal is less than 0.01?

  For example,.

  Counter.Count = signal<=>

  The total number of rain is

  Counter * 10 + signal

  Use TimeOfDay or TimeOfWeek to reset the counter.

• Set the calendar to display day, month and year

  In addition to the time of day, my computer began giving the day of the week (Monday) only instead of day, month and year as it did before. Tried to go back to 29/12/2014 or something similar without success.  Any suggestions?

  Hello

  Sorry for the late reply and I appreciate your patience.

  User profile corruption might have caused the problem. As a solution, I suggest to create a new user profile and check if the date format displays the same thing.

  To do this, see the link below.

  http://Windows.Microsoft.com/en-us/Windows/fix-corrupted-user-profile#1TC=Windows-7.

  It will be useful. Just answer us with the State, and we will be happy to help you.

• The date system in the taskbar indicates the year, month, day (as in the army). How can I read the day, month, year?

  The date system in the taskbar indicates the year, month, day (as in the army). How can I read the day, month, year?

  It's weird, in my version (Windows 7 Pro), it shows much more than that.

• split with day, month and year in a date

  Hi all

  I have this point of view

  with (id, dt) t (select 1, to_date('02/06/2010','MM/DD/YYYY') of all the double union)
  Select 2, to_date('11/29/2001','MM/DD/YYYY') of all the double union
  Select 3, to_date('02/06/2011','MM/DD/YYYY') of all the double union
  Select option 4, double to_date('10/10/2011','MM/DD/YYYY')
  )
  --
  -end of test data
  --
  SELECT id, dt
  floor (months_between(sysdate,dt)/12) years
  , floor (mod (months_between (sysdate, dt), 12)) months
  -The day is ambiguous because of the number of days in a different month
  -you could design your own algorithm to give the desired results, but it is a good approximation
  , floor (sysdate-(add_months (dt, floor (months_between (sysdate, dt))) as dys
  t
  /
  ID DT YEARS MONTHS DYS
  ---------- ----------- ---------- ---------- ----------
  1-6 FEBRUARY 2010 5 8 22
  2 29 NOVEMBER 2001 13 10 29
  3-6 FEBRUARY 2011 4 8 22
  4-10 OCTOBER 2011 4 0 18

  As a result, 26 years old, 26 months, 91 days.
  

  
  

  And subtract years, MONTHS and DYS as format

  erase years, from 1 to 12 months and the days of the 1 to 30 as

  91 days = 30 x 3 + 1 = > 1 day over 3 months

  26 months + 3 months = 29 months = > 24 months + 5 months = 2 years + 5 months

  and 26 years + 2 years = 28

  desire to result: 28 years, 5 months and 1 day.

  Is there a function that give me this result from the query above?

  Kind regards

  Gordan

  Hello Gordan,.

  I understand that you are the SUM of all the lines.

  Here are two ideas:

  -1 - using a date
  (a) SUM (TRUNC (sysdate) - dt)
  I would give you the total number of days.
  (b) adding the number of days to, for example, DATE ' 1900-01-01' would give another date...
  (c) simply subtract the years 1900 and it takes several years,
  subtract 1 from the month (1-12 from January to December) and you have the number of months
  subtract 1 from the day of the month (1-31) and you have the number of days.
  It is of course 'approximate' as the months do not have the same number of days...

  -2 - using only the number of days
  Maybe you want something else, for example: the total number of days divided by 365.25 for many years, the recall divided by 30 to get the number of months, the reminder being the number of days.

  (a) SUM (TRUNC (sysdate) - dt)

  I would give you the total number of days.

  (b) Division by 365.25 (le.25 is to take leap years into account, but you can choose for example "365")

  (c) number of reminder of days (total - years * 365.25): divided by 30 is the number of months

  (d) number reminder of days (total - years * 365.25 - months * 30): gives the number of days.

  The following two options with your test data (by chance, this gives the same result in this case)

  option 1:
  with (id, dt) t (select 1, to_date('02/06/2010','MM/DD/YYYY') of all the double union)
  Select 2, to_date('11/29/2001','MM/DD/YYYY') of all the double union
  Select 3, to_date('02/06/2011','MM/DD/YYYY') of all the double union
  Select option 4, double to_date('10/10/2011','MM/DD/YYYY')
  )
  target_date AS
  (SELECT DATE ' 1900-01-01' + SUM (TRUNC (sysdate) - t.dt) FROM t trgt)
  in_pieces AS
  (SELECT TO_NUMBER (TO_CHAR (td.trgt, 'YYYY')) - 1900 y
  , TO_NUMBER (TO_CHAR (td.trgt, 'MM')) - 1 m
  , TO_NUMBER (TO_CHAR (td.trgt, 'DD')) - 1 d
  Target_date TD
  )
  SELECT "result: ' |"
  TRIM (CASE WHEN ip.y = 0 THEN NULL
  WHEN ip.y = 1 THEN "1 year"
  Of OTHER TO_CHAR (ip.y) | "years".
  END |
  CASE WHEN ip.m = 0 THEN NULL
  WHEN ip.m = 1 THEN "1 month"
  Of OTHER TO_CHAR (ip.m) | 'months '.
  END |
  CASE WHEN ip.d = 0 THEN NULL
  WHEN ip.d = 1 THEN "1 day"
  Of OTHER TO_CHAR (ip.d) | 'days '.
  END
  )
  Of in_pieces ip
  ;
  result: 28 years, 4 months and 28 days

  option 2:

  with (id, dt) t (select 1, to_date('02/06/2010','MM/DD/YYYY') of all the double union)
  Select 2, to_date('11/29/2001','MM/DD/YYYY') of all the double union
  Select 3, to_date('02/06/2011','MM/DD/YYYY') of all the double union
  Select option 4, double to_date('10/10/2011','MM/DD/YYYY')
  )
  nb_days AS
  (SELECT SUM (TRUNC (sysdate) - t.dt) n t)
  , y AS
  (SELECT Nb.n, FLOOR (nb.n / 365.25) y nb_days n. b.)
  ym AS
  (SELECT y.n, y.y, FLOOR ((y.n-y.y * 365,25) / 30) FROM a)
  ymd AS
  (SELECT ym.y, ym.m, ym.n - ym.y * 365.25 - ym.m * ym FROM 30 d)
  SELECT "result: ' |"
  TRIM (CASE WHEN ymd.y = 0 THEN NULL
  WHEN ymd.y = 1 THEN "1 year"
  Of OTHER TO_CHAR (ymd.y) | "years".
  END |
  CASE WHEN ymd.m = 0 THEN NULL
  WHEN ymd.m = 1 THEN "1 month"
  Of OTHER TO_CHAR (ymd.m) | 'months '.
  END |
  CASE WHEN ymd.d = 0 THEN NULL
  WHEN ymd.d = 1 THEN "1 day"
  Of OTHER TO_CHAR (ymd.d) | 'days '.
  END
  )
  WAN
  ;
  result: 28 years, 4 months and 28 days

  Best regards

  Bruno Vroman.

• difference between two dates in the year/month/day in Obiee

  Hi gurus,

  I know this question have been asked and answered several times but I have a requirement that is a little different, then the previous ones.

  I want to calculate the difference between two dates in OBIEE10g in year/month/day format similar to the below SQL output

  SQL > select end_date, start_date,

  trunc (months_between (end_date, start_date) / 12) years.

  months of mod (trunc (months_between (end_date, start_date)), 12).

  End_date - add_months (start_date, trunc (months_between (end_date, start_date))) days

  t

  Thanks in advance

  SK

  Search for this

  "TIMESTAMPDIFF IN THE FORM OF MTHS # YEARS."

  or else

  To get the current mandate of employees since the date of hiring in the form of # years # mths or # year (s) # month (s)

  Concat (concat (cast (TIMESTAMPDIFF (SQL_TSI_Month, "Employee attributes". (((' ' Hiring last Date employee ", CURRENT_DATE) / 12 as char),"Year (s)"), concat (cast (MOD (TIMESTAMPDIFF (SQL_TSI_Month,"Employee attributes". (((((' ' Hiring last Date employee ", CURRENT_DATE), 12) as char), 'Month (s)'))

  Concat (concat (CAST (TIMESTAMPDIFF (SQL_TSI_YEAR, "Employee attributes". (((' ' Hiring last Date employee ", CURRENT_DATE) as CHAR),"Year (s)"), concat (cast (MOD (TIMESTAMPDIFF (SQL_TSI_Month,"Employee attributes". (((((' ' Hiring last Date employee ", CURRENT_DATE), 12) as char), 'Mth (s)'))

• Get the number of days in a month based on the month and year of fields

  I have a column in my form which lists the days in a month. I want to configure a hidden field that calculates the total number of days in a month, based on the month and year of the field inputs. The number of days will determine what appears on the column. For example, if I put 4 months, and 2016 in the field of the year, I get 30 in the hidden field. Thus, on the column 'Day', I'll have numbers 1-30. Or if I put 2 months and 2016 in the field of the year, I get the 29 in the hidden field. If the numbers 1-29 appears in the column 'day '.

  Found this on some forum javascript code:

  //Month is 1 based
function daysInMonth(month,year) {
   return new Date(year, month, 0).getDate();
}

//July
daysInMonth(7,2009); //31
//February
daysInMonth(2,2009); //28
daysInMonth(2,2008); //29

  I do not know how to convert this code in JavaScript to adobe and don't really know how to use it. All I know how to do is to configure the field values for the field month and year as variables. I am a novice programmer and would appreciate it really all the help I can get. Thank you in advance!

  The code seems to be JavaScript and runs as needed by using the JavaScript console.

  I would like to consider making more general code, so if you have a date string that includes at least the month and year we could just call the function and get the number of days for that month.

  The following script will calculate the number of days in a month, by using at least the month and year values can display the result on the JavaScript console and all of the value field for the field that has this code as the custom calculation Script.

  function daysInMonth (oDate) {}
  return new Date (oDate.getFullYear (), oDate.getMonth () + 1, 0) .getDate ();
  }

  nMonth var = this.getField("Month").valueAsString; get the value of month;
  nYear var = this.getField("Year").valueAsString; get the value of the year;

  Event.Value = "";

  If (nMonth! = "" & nYear!) = "") {}
  var MyDate = util.scand ("' / mm/yyyy ', nMonth +" / "+ nYear); convert to date object;
  var nDaysInMonth = daysInMonth (MyDate); get the number of days;

  Console.Open (); Open the JavaScript console;

  Console.clear(); clear the console;

  Console.println ("Days in" + nMonth + ":" + nDaysInMonth); show days in month;

  Event.Value = nDaysInMonth; Set the value of the field;

  }

• Suggest the Sql query to get the years, months, days of the two dates.

  Hi all

  SELECT TO_DATE('06-02-2014','DD-MM-YYYY') - TO_DATE('01-04-1998','DD-MM-YYYY') SERVICE_DAYS OF THE DOUBLE

  5790 SERVICE_DAYS

  POWER REQUIRED:
  YEARS MONTHS DAYS
  15 10 5

  OUTPUT DISPLAY:

  Select
  Floor(5790/ 365.25),
  Floor (mod (5790,365.25) / (365.25 / 12)),.
  ceil (mod(5790,365.25 / 12))
  Double;

  YEARS MONTHS DAYS
  15 10 7 - TWO DAYS MORE

  
  How they calculate manually is as follows:

  2014 02-06
  1998-04-01
  ---------------
  05 - days
  
  2014 02-06
  1998-04-01
  + 12
  ---------------
  10 month 02 is lower than 04, so we borrow 12 months of TI (2014) years will be (2013)
  
  2014 02-06
  1998-04-01
  -(1).
  ---------------
  15. for the years since we-1

  Kind regards

  Afzal.

  OK... so try not not to multiply the reminder to add_months with 30 but to solve the 'day-thing' in a different way...

  first step (data) is only my implementation testdata.

  WITH DATA AS (SELECT TO_DATE('02-03-2014','DD-MM-YYYY') AS Degnide

  TO_DATE('07-04-2008','DD-MM-YYYY') AS dayfrom

  OF the double

  UNION ALL

  SELECT TO_DATE('12-04-2014','DD-MM-YYYY') AS Degnide

  TO_DATE('11-05-1994','DD-MM-YYYY') AS dayfrom

  OF the double

  UNION ALL

  SELECT TO_DATE('06-02-2014','DD-MM-YYYY') AS Degnide

  TO_DATE('01-04-1998','DD-MM-YYYY') AS dayfrom

  OF the double

  )

  calc AS (SELECT months_between (dayto, dayfrom) SERVICE_DAYS

  Degnide

  dayfrom

  DATA)

  SELECT Degnide

  dayfrom

  floor(service_days/12) IN years

  floor (mod(service_days,12)) AS month

  , BUSINESS sign (EXTRACT (Degnide DAY) - EXTRACT (DAY FROM dayfrom))

  WHEN THEN-1 EXTRACT (DAY FROM Degnide) + 30 - EXTRACT (DAY OF THE dayfrom)

  WHEN 0 then 0

  WHEN 1 THEN EXTRACT (Degnide DAY) - EXTRACT (DAY FROM dayfrom)

  0 OTHERWISE

  end as days

  CALC

  HTH

• Get the number of days, months and year

  Hi all
  
  I'm using OBIEE 11 g. I have a line of dash for year (2010, 2011, etc.) and months (Jan, Feb,..., Dec). How can I get the number of days in a given month and year in my report?
  
  For example, if I choose to Jan & 2011, I'd get 31 and 356. And if I choose Feb & 2011, I would get 28 and 356.
  
  Thank you!

  I have something more good option here... Try to use this

  DAYOFMONTH (TIMESTAMPADD (SQL_TSI_DAY, DAYOFMONTH (CURRENT_DATE) *-1, TIMESTAMPADD (SQL_TSI_MONTH, 1, CURRENT_DATE)))

  This will give you the days for the month in progress (Oct 2011) after calculating the last day of each month... Similary dynamically pass the value Date or month... You will need to modify this function but shud be simple. Hope this helps :)

• 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.

• Select SQL - last day of the month, a year from now

  Hello
  How do we get the last day of the month, a year from now.
  
  For example-sysdate-12/2/2009
  Result - 31/12/2009
  
  Thank you

  Hello

  SELECT  LAST_DAY ( ADD_MONTHS ( SYSDATE
                                , 12
              )           )
  FROM    dual;
  

  Remember, all DATEs, including the results of this query, include the hours minutes and seconds.
  The hours, the minutes and the seconds returned by the above expression are the same as SYSDATE: If you run it at 15:43:30 today, it returns 15:43:30 on December 31, 2010. This can be important if you want to use an expression like this as the cut-off point in a WHERE clause.