extraction of data each month and year wise in a table

Similar Questions

• problem to insert only month and year instead of the full date

  select b.penjara_id, p.penj_lokasi, a.no_daftar, b.episod, b.nama1,to_char(a.trkh_mula_prl,'dd/mm/yyyy') as trkh_mula_prl, to_char(bulan_proses,'mm/yyyy') as bulan_proses,
          b.epd, b.lpd
  from prl_daftar_proses a, senarai_pesalah b, penjara p
  
  where a.no_daftar=b.no_daftar
  and a.episod=b.episod
  and b.penjara_id = p.penjara_id
  and a.setuju_jplp is null
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  order by b.penjara_id, a.bulan_proses,a.no_daftar

  Hi, can someone help me how I can register only the month and year of the value which have full day in the database?
  
  for example, the date is 18/09/2012, but I want to just insert 09/2012 as a parameter. If I want to insert only one setting, I can do... But I have problem when I want to insert two parameters...

  Lina saleh wrote:
  I already try before that and he invites me error like this...
  "name of the invalid host/identification variable.

  Change the name of the connection variable to: from_date and: to_date

  Is a reserved keyword.

• get the month and year of date type

  Hi all
  I need to get the month and year of type date.
  
  For example

  select to_date('2011-01-17', 'yyyy-mm-dd') from dual;

  Required result:

  01-2011

  Any ideas?
  
  Thanks in advance,
  Bahchevanov.

  select to_char(to_date('2011-01-17', 'yyyy-mm-dd'),'mm-yyyy') from dual;
  

  HTH

• 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.

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  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<=>

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  Counter * 10 + signal

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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.
  
  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 :)

• spend months and year in a query

  I created_date in my table... I want to only get the records of a month or a year of records
  
  How we can write a query to spend months and year to a query
  
  example is my name of the x table
  
  Select * from x where trunc (created_date) =?

  Hello

  Rather than convert all DATES in your array to a string and then compare these values to a parameter of sting, it would be more efficient to convert your DATE string parameter and then compare all your DATEs unchanged to her.
  For example, if: year_month is the string "201108', you might say:

  SELECT  *
  FROM    x
  WHERE   created_date >=             TO_DATE (:year_month || '01', 'YYYYMMDD')
  AND     created_date < ADD_MONTHS ( TO_DATE (:year_month || '01', 'YYYYMMDD')
                        , 1
                        )
  ;
  

  I hope that answers your question.
  If not, post a small example of data (CREATE TABLE and INSERT statements) for table x, some values for: year_month, and the desired results for each value of: year_month from these data.
  Always tell what version of Oracle you are using.

• 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.

• 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;

  }

• to get the month and year separated

  Hello
  
  I have the entry_date_fin of column (type of data-varchar2 (8))
  
  with values like Apr.-2008
  Feb-2009 etc.
  
  can I get the distinct month and year separated

  i tried select to_date(entry_date_fin,'MM') FROM  xyz  doesnt wrk

  kindly guide me

  Although you are (wrongly, IMO) storage of dates as strings, just use string manipulation...

  SQL> with t as (select 'FEB-2008' as txt from dual)
    2  --
    3  select substr(txt,1,3) as mnth
    4        ,substr(txt,5) as yr
    5  from t
    6  /
  
  MNT YR
  --- ----
  FEB 2008
  
  SQL>
  

• 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.

• Calculate the age (months and years) of two dates

  Trying to get the (almost exact) age in years and months to fill in 'Age at the time of retirement' (years1) since the date of birth (DOB1) and the date of their departure in retirement (retireDate1). The only examples that I could find are dates "today." If anyone can help with code? And I understand it works better with JavaScript but FormCalc would also work?

  Dates as DD/MM/YYYY

  Field is num {z9}

  Thank you in advance for your help!

  I could do this work with Age1 as a text field. Of course, you can do whatever you want with it, but this method makes the most sense for me because I can say 99 months "99 years". If you need to separate the year and month in different areas, you should be able to do it easily. Finally, it requires that the user does not manually enter dates. (Something on the raw values isn't quite right when they came from the selection.)

  Age1::calculate (or anywhere where you want this event)

  if (!DOB1.isNull && !retireDate1.isNull){
    var d1 = DOB1.rawValue.split('-');
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• ATTN: Staff [added kglad] double load for each month and one subscription

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

  Hello

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  http://www.Adobe.com/2006/mxml '.

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  [Bindable]

  private var actualYear:String;

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• Month and year Validation

  Hi all!
  
  I want to check if a given date in my form is lower or higher than sysdate...
  
  I used the coding

  if trim(:empaply.fromdate)<trim(sysdate) then
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  This condition verifies that the date and not the month and the year...
  
  So, how can we check the month and the year is less than the current sysdate in sql?
  
  pls help me with a solution...
  
  Thanks and greetings
  user 10685325

  "Assuming that" fromdate "is the date data type, then it will simply be.

  if :empaply.fromdate < sysdate then
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• How to get the previous month and year accordingly

  Hi friends,
  I wanted to select the month as a number like 4 and the new year 2011
  but I want to choose the previous month from sysdate
  as if the sysdate is June 10, 2011
  It should give me 5 months and
  in the year, it should give as 2011
  If the sysdate is January 1, 2012
  the month should be like 12 and year should be in 2011
  
  Thank you

  Hello

  776317 wrote:
  Thank you very much I just the 0 being the prefix... can you pls how to avoid that too well pls

  TO_CHAR ( ADD_MONTHS (SYSDATE, -1)
          , 'FMmm yyyy'
         )
  

  For more information on the FM, wee manual SQL language modifier works:
  http://download.Oracle.com/docs/CD/B28359_01/server.111/b28286/sql_elements004.htm#sthref456