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

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  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.

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

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

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• Manage the values of the actual day, month and year

  Hello

  I am trying to enter digital variables the same day (format: 1 / 10), month (format: 1 / 10) and year (format: 1900). I tried to use this sample:

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  layout = "absolute" creationComplete = "onCreationComplete ()" > "

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• 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.

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• How to get the previous month and year accordingly

  Hi friends,
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• 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:

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function daysInMonth(month,year) {
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}

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

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  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.

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  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!) = "") {}
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  Console.Open (); Open the JavaScript console;

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  }