BEEP Scheduler - last day of the month

Similar Questions

• Get the last day of the month

  Hello

  How can I get the (day number of the) last day of a given month and year?

  So, if

  year 2015 = and month = 1, the output should be 31

  year = 2012 and month = 2 while production is expected to be 29

  year 2013 = and month = 2 while production is expected to be 28

  etc.

  Y at - it no OBIEE to achieve? If not, then a command SQL Oracle is very good too (I prefer OBIEE however)

  Thanks in advance,

  Erik

  You start from the first day of the month, you add 1 month [TIMESTAMPADD (SQL_TSI_MONTH, 1, 'your date')], you must subtract 1 day [TIMESTAMPADD (SQL_TSI_DAY,-1, 'your date')] and use the DAYOFMONTH function to get the number of the last day of the month [DAYOFMONTH ('your ' date')].

  All these calls nesting together and you're there.

• Breaking a year in coming weeks of beginning and end Dates (parameters) and first and last day of the month

  Hello people:

  I have currently a query that picks up all the weeks between a date range where my week starts on Thursday and ends on Wednesday. The following query works fine except that I need the week of beginning and end, from beginning and end Dates of months and the beginning and the Dates of end of the values of the settings, I'm passing. In this case, I chose January 1, 2013 to 31 December 2013.

  Any help or pointers would be great!

  Thank you!

  Problem: I am picking up days of December 2012 as the first week began on December 27, 2012. In addition, in December, I'm pretty much lost last week as well (26 December - 31 December).

  The request in hand:

  SELECT first_thursday + (7 * (LEVEL - 1))   AS week_start_date,
                   first_thursday + (7 * LEVEL) - 1    AS  week_end_date
  FROM
  (
    SELECT TRUNC(p_from_date + 4, 'IW') - 4   AS first_thursday, -- Week should start the pre ceeding THURSDAY based on the Start Date
                    TRUNC( p_to_date + 4,  'IW') - 5     AS last_wednesday
    FROM
    (
      SELECT to_date('01-JAN-2013') AS p_from_date,
                      NVL(to_date('31-DEC-2013'), SYSDATE) AS p_to_date
      FROM     dual
    ) parms
  ) end_points
  CONNECT BY LEVEL <= ( last_wednesday + 1 - first_thursday)/7;
  

  Currently, this is the result I get (I'm only including the months of January and December here).

  Week_Start_Date     Week_End_Date
  27-DEC-12                02-JAN-13
  03-JAN-13                 09-JAN-13
  10-JAN-13                 16-JAN-13
  17-JAN-13                 23-JAN-13
  24-JAN-13                 30-JAN-13
  31-JAN-13                 06-FEB-13
  

  December:

  28-NOV-13               04-DEC-13
  05-DEC-13              11-DEC-13
  12-DEC-13              18-DEC-13
  19-DEC-13              25-DEC-13
  

  What I would really like is based on start and end Dates, as well as the first and the last day of the month is similarly try nicely. I have provided and then gently break at the end of the month:

  January:

  01-JAN-13              02-JAN-13
  03-JAN-13              09-JAN-13
  10-JAN-13             16-JAN-13
  17-JAN-13             23-JAN-13
  24-JAN-13             30-JAN-13
  31-JAN-13             31-JAN-13
  

  February:

  01-FEB-13    06-FEB-13
  07-FEB-13    13-FEB-13
  14-FEB-13    20-FEB-13
  21-FEB-13    27-FEB-13
  28-FEB-13    28-FEB-13
  

  November:

  31-OCT-13    06-NOV-13
  07-NOV-13    13-NOV-13
  14-NOV-13    20-NOV-13
  21-NOV-13    27-NOV-13
  28-NOV-13    04-DEC-13
  

  December:

  01-DEC-13           04-DEC-13
  05-DEC-13           11-DEC-13
  12-DEC-13          18-DEC-13
  19-DEC-13          25-DEC-13
  26-DEC-13          31-DEC-13
  

  Hello

  Roxyrollers wrote:

  Hello people:

  I have currently a query that picks up all the weeks between a date range where my week starts on Thursday and ends on Wednesday. The following query works fine except that I need the week of beginning and end, from beginning and end Dates of months and the beginning and the Dates of end of the values of the settings, I'm passing. In this case, I chose January 1, 2013 to 31 December 2013.

  Any help or pointers would be great!

  Thank you!

  Problem: I am picking up days of December 2012 as the first week began on December 27, 2012. In addition, in December, I'm pretty much lost last week as well (26 December - 31 December).

  The request in hand:

  1. First_thursday SELECT + (7 * (LEVEL - 1)) AS week_start_date,
  2. first_thursday + (7 * LEVEL)-1 AS week_end_date
  3. Of
  4. (
  5. SELECT TRUNC (p_from_date + 4, 'IW') - 4 AS first_thursday,-week should start to perform the pre-mounted THURSDAY based on the Start Date
  6. TRUNC (p_to_date + 4, 'IW') - 5 AS last_wednesday
  7. Of
  8. (
  9. SELECT to_date('01-JAN-2013') AS p_from_date,
  10. NVL (to_date('31-Dec-2013'), SYSDATE) AS p_to_date
  11. OF the double
  12. ) parms
  13. ) end_points
  14. CONNECT BY LEVEL<= (="" last_wednesday="" +="" 1="" -="">

  Currently, this is the result I get (I'm only including the months of January and December here).

  1. Week_Start_Date Week_End_Date
  2. DECEMBER 27, 12 2 JANUARY 13
  3. JANUARY 3, 13 JANUARY 9, 13
  4. 10 JANUARY 13 JANUARY 16, 13
  5. 17 JANUARY 13 23 JANUARY 13
  6. 24 JANUARY 13 30 JANUARY 13
  7. 31 JANUARY 13 FEBRUARY 6, 13

  December:

  1. NOVEMBER 28, 13 4 DECEMBER 13
  2. 5 DECEMBER 13 DECEMBER 11, 13
  3. 12 DECEMBER 13 18 DECEMBER 13
  4. 19 DECEMBER 13 DECEMBER 25, 13

  What I would really like is based on start and end Dates, as well as the first and the last day of the month is similarly try nicely. I have provided and then gently break at the end of the month:

  January:

  1. JANUARY 1, 13 2 JANUARY 13
  2. JANUARY 3, 13 JANUARY 9, 13
  3. 10 JANUARY 13 JANUARY 16, 13
  4. 17 JANUARY 13 23 JANUARY 13
  5. 24 JANUARY 13 30 JANUARY 13
  6. 31 JANUARY 13 JANUARY 31, 13

  February:

  1. 1ST FEBRUARY 13 FEBRUARY 6, 13
  2. 7 FEBRUARY 13 FEBRUARY 13, 13
  3. 14 FEBRUARY 13 FEBRUARY 20, 13
  4. 21 FEBRUARY 13 FEBRUARY 27, 13
  5. 28 FEBRUARY 13 FEBRUARY 28, 13

  November:

  1. 31 OCTOBER 13 NOVEMBER 6, 13
  2. 7 NOVEMBER 13 NOVEMBER 13, 13
  3. 14 NOVEMBER 13 NOVEMBER 20, 13
  4. 21 NOVEMBER 13 NOVEMBER 27, 13
  5. NOVEMBER 28, 13 4 DECEMBER 13

  December:

  1. 1ST DECEMBER 13 DECEMBER 4, 13
  2. 5 DECEMBER 13 DECEMBER 11, 13
  3. 12 DECEMBER 13 18 DECEMBER 13
  4. 19 DECEMBER 13 DECEMBER 25, 13
  5. 26 DECEMBER 13 DECEMBER 31, 13

  Why the "weeks" in November, have all 7 days, while the "weeks" in the first or the last week in the other month usually have less?  (For example, February 28 is a 'week' alone.)

  The following works for the other months:

  WITH all_days AS

  (

  SELECT DATE "2013-01-01' + LEVEL - AS a_date 1

  OF the double

  CONNECT BY LEVEL<=>

  )

  SELECT DISTINCT

  More GRAND (TRUNC (a_date + 4, 'IW') - 4)

  , TRUNC (a_date, 'MONTH')

  ) AS week_begin

  , The LEAST (TRUNC (a_date + 4, 'IW') + 2

  TRUNC (LAST_DAY (a_date))

  ), Week_end

  Of all_days

  ORDER BY week_begin

  ;

  Output:

  WEEK_BEGIN WEEK_END

  ----------- -----------

  January 1, 2013 January 2, 2013

  January 3, 2013 January 9, 2013

  January 10, 2013 January 16, 2013

  January 17, 2013 January 23, 2013

  January 24, 2013 January 30, 2013

  January 31, 2013 January 31, 2013

  February 1, 2013 February 6, 2013

  February 7, 2013 February 13, 2013

  February 14, 2013 February 20, 2013

  February 21, 2013 February 27, 2013

  February 28, 2013 February 28, 2013

  ...

  November 1, 2013 November 6, 2013

  November 7, 2013 November 13, 2013

  November 14, 2013 November 20, 2013

  November 21, 2013 November 27, 2013

  November 28, 2013 November 30, 2013

  December 1, 2013 December 4, 2013

  December 5, 2013 December 11, 2013

  December 12, 2013 December 18, 2013

  December 19, 2013 December 25, 2013

  26 December 2013 31 December 2013

• Last day of the month with the last minute of the day

  Hello
  I wonder about how to get the last day of the month with the last minute of this day(23:59). We can use the last_day function (DATE) to get the last date, but reminds me of the time of day when the function has been executed. But I want to have the last minute with my last date of the current month.
  
  Please inform.
  
  Thank you
  
  HP

  Hello

  Ora_bie wrote:
  Hello
  I wonder about how to get the last day of the month with the last minute of this day(23:59). We can use the last_day function (DATE) to get the last date, but reminds me of the time of day when the function has been executed. But I want to have the last minute with my last date of the current month.

  In fact, it returns at the same time as the argument you pass; But whatever it is, is not what you want.

  This will return the last day of the month that contains the DATE 23:59 (or 23:59) dt:

  SELECT  ADD_MONTHS ( TRUNC (dt, 'MONTH')
               , 1
               ) - (1 / (24 * 60))
  FROM    dual
  ;
  

  TRUNC (dt, 'MONTH') is 00:00 on the first day of the month that contains dt.
  ADD_MONTHS (TRUNC (dt, 'MONTH'), 1) is from 00:00 the first day of the month following the one containing dt.
  ADD_MONTHS (TRUNC (dt, 'MONTH'), 1)-(1 / (24 * 60)) is a minute earlier.
  (Since there are 24 hours, each with 60 minutes, every day, 1 / (24 * 60) days is identical to a minute.)

  Another (less clear, in my opinion) way to get the same results is:

  TRUNC (LAST_DAY (dt)) + (1439/1440)  -- 1440 minutes = 1 day
  

• 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.

• ORA-01847: day of the month must be between 1 and the last day of the month

  Hi all
  
  
  
  
  
  
  Select
  To_date(mDate||) e '|| mtime, "YYYY-MM-DD HH24MISS") meddate
  of PR_MOCTEST_TEMP
  
  mDate IS OF TYPE of DATA VARCHAR2 AND mtime OF VARCHAR2
  
  mDate: 07-09-2009
  mtime: 131241
  
  mDate IS IN FORMAT DD-MONTH-YYYY
  mtime: HHMMS
  
  mtimeOF IS OF TYPE of DATA VARCHAR2 AND VARCHAR2 mDate
  
  mDate: 07-09-2009
  mTIME: 131241
  
  mDATE IS IN FORMAT DD-MONTH-YYYY
  mTIME: HHMMS
  
  I NEED TO COMBINE THE 2 COLUMNS TO GET THE OUTPUT AS DD-MM-YYYY HH24MISS
  AND I AM GETTING THE ERROR;
  
  
  ORA-01847: day of the month must be between 1 and the last day of the month
  
  where mistaken?
  
  Please notify
  
  
  Kai

  You will find the bad dates using a table of logging of errors:

  --we'll try to insert into here, invalid dates will be thrown out
  create table dst(d date);
  
  -- create an error log table dst_err
  exec dbms_errlog.create_error_log('dst', 'dst_err');
  
  -- try the insert. Will fail with your ORA-01847
  insert into dst select TO_DATE(mdate||' '||mtime,'DD-MM-YYYY HH24MISS') from PR_MOCTEST_TEMP log errors into dst_err('comment'); 
  
  -- view the errors
  select * from dst_err;
  

• Get the first and the last day of the month...

  Hi friends,
  
  I'm trying to get the first and last day of a month and wants to implement that in the following code:
  
  SELECT COALESCE (Date_A, Date_B, Date_C)
  OF the double
  
  Here Date A and B are in Format DD MM YYYY (March 14, 2008)
  and Date like MM YYYY (March 2008)
  
  How can I get the Date_C as of March 1, 2008, OR March 31, 2008 if Date A and B are NULL?
  
  Thank you!
  
  Published by: user11095386 on April 23, 2009 10:45

  Hello

  In my previous message, I thought that you were starting with strings like ' 03 12 2009"and you want to display in the form of" 12 March 2009'.» If what you have is just the opposite, then simply reverse strings format in my first message. Add a comma, if you want, in the format string.
  I think that's what you want:

  COALESCE ( TO_CHAR ( TO_DATE ( Date_A, 'fmMonth DD YYYY'), 'MM DD YYYY')
           , TO_CHAR ( TO_DATE ( Date_B, 'fmMonth DD YYYY'), 'MM DD YYYY')
           , TO_CHAR ( TO_DATE ( Date_C, 'fmMonth YYYY'),    'MM DD YYYY')
           )
  

  Notice how, on the 3rd line, TO_DATE is called without DD in the format string:

  TO_DATE ( Date_C, 'fmMonth YYYY')
  

  When you do so, by default the day the 1st of the month, so that's all you need to do to convert the VARCHAR2 'March 2009' at the DATE of March 1, 2009.

  If you would like the last day of the month, not the first, when Date_C is selected, then use LAST_DAY:

  COALESCE ( TO_CHAR ( TO_DATE ( Date_A, 'fmMonth DD YYYY'), 'MM DD YYYY')
           , TO_CHAR ( TO_DATE ( Date_B, 'fmMonth DD YYYY'), 'MM DD YYYY')
           , TO_CHAR ( LAST_DAY ( TO_DATE ( Date_C
                                            , 'fmMonth YYYY'
                              )
                       )
               , 'MM DD YYYY'
               )
           )
  

• First day/last day of each month

  Hello
  y at - it sql/script to get the first day/last day of each month.new to bi do not know the syntax
  
  If a point on the link to get these things will be highly appreciated

  Hello
  I suggest that you go through the following link

  (1) http://oracle.ittoolbox.com/groups/technical-functional/oracle-bi-l/obiee-date-comparison-1726677#M1729534

  Toget first/last day of the month

  (2) http://www.obinotes.com/2010/02/tip-to-get-firstday-lastday.html

  New to OBIEE---> http://obiee101.blogspot.com/2009/07/obiee-how-to-get-started.html

  Thank you
  Saichand.V

• How to get sales last day of the previous month

  HELO guys

  I have a question how do I get sales date last day of the previous month where my fact table is level. in fact I tried using Ago but it seems not working. for example, if I chose December 29, 2014 so if I used the function there are then the result will display data on November 29, 2014 (not the last day of November 2014).

  Please guys if you have any solution and I really advise thanks to you.

  Not exactly under what circumstances specific as you try to reach this goal, but suppose you have a few guest dashboard where you select a day and you have a report with a measure where you want to show the sales of the last day of the previous month based on the selected date in the dash prompt.

  If you change the formula of this measure to something like:

  FILTER (with the HELP of 'Sales' (DATE = TIMESTAMPADD (SQL_TSI_DAY-1, TIMESTAMPADD (SQL_TSI_DAY, DAYOFMONTH(@{PV_DATE_SELECTED}) * (-1) + 1, @{PV_DATE_SELECTED}))))

• How to get the last day of the weekend rather than a month

  Hi all

  Is the request appropriate and preferable to use to get the last business day of the month below?

  SELECT to_date('13.08.2014','dd.mm.yyyy') + MAX(RNUM) LastDay(nonweekend)
     FROM   (SELECT ROWNUM RNUM
            FROM   ALL_OBJECTS where rownum <=31)
    WHERE   ROWNUM <= 31 ANd
       TO_CHAR(to_date('13.08.2014','dd.mm.yyyy') + RNUM, 'DY','NLS_DATE_LANGUAGE=ENGLISH' ) NOT IN ('SAT' , 'SUN')
       and to_date('13.08.2014','dd.mm.yyyy') + RNUM <= last_day(to_date('13.08.2014','dd.mm.yyyy'))
  

  I have to use this motion for production.

  East - recommended to use?

  You can change your query this way:

  WITH t AS (SELECT trunc (to_date ('& calc_date', 'dd.mm.yyyy'), 'mm') + level - 1 calc_date)

  OF THE DOUBLE

  CONNECT BY LEVEL<= last_day(to_date('&calc_date',="" 'dd.mm.yyyy'))="" -="" trunc(to_date('&calc_date',="" 'dd.mm.yyyy'),="" 'mm')="" +="">

  )

  Select max (calc_date)

  t

  where to_char (calc_date, 'DY', 'NLS_DATE_LANGUAGE = ENGLISH') NOT IN ("SAT", "Sun");

• How to get the last day of a month for every 2 months for a given period?

  Hello

  Can is it some please let me know how to get the last day, last day of the week, the weekend last day, last Monday, one month for every 2 months for a given period?

  Thanks in advance.

  Try the below

  SELECT LAST_DAY (ADD_MONTHS (TRUNC(startdate,'MM'), ((LEVEL*2)-2))) lastday.

  BOX WHEN TO_CHAR (LAST_DAY (ADD_MONTHS (TRUNC(startdate,'MM'), ((LEVEL*2)-2))), 'DY') = 'SAT '.

  SO LAST_DAY (ADD_MONTHS (TRUNC(startdate,'MM'), ((LEVEL*2)-2)))-1

  WHERE TO_CHAR (LAST_DAY (ADD_MONTHS (TRUNC(startdate,'MM'), ((LEVEL*2)-2))), 'DY') = 'Sun '.

  THEN LAST_DAY (ADD_MONTHS (TRUNC(startdate,'MM'), ((LEVEL*2)-2)))-2

  Of OTHER LAST_DAY (ADD_MONTHS (TRUNC(startdate,'MM'), ((LEVEL*2)-2)))

  END as lastweekday,

  BOX WHEN TO_CHAR (LAST_DAY (ADD_MONTHS (TRUNC(startdate,'MM'), ((LEVEL*2)-2))), 'DY') IN ('Sam', 'SUN')

  THEN LAST_DAY (ADD_MONTHS (TRUNC(startdate,'MM'), ((LEVEL*2)-2)))

  Of OTHER LAST_DAY (ADD_MONTHS (TRUNC(startdate,'MM'), ((LEVEL*2)-2)))

  -(TO_NUMBER (TO_CHAR (LAST_DAY (ADD_MONTHS (TRUNC(startdate,'MM'), ((LEVEL*2)-2))),' from)) - 1).

  END as lastweekendday,

  BOX WHEN TO_CHAR (LAST_DAY (ADD_MONTHS (TRUNC(startdate,'MM'), ((LEVEL*2)-2))), 'DY') = 'MY

  THEN LAST_DAY (ADD_MONTHS (TRUNC(startdate,'MM'), ((LEVEL*2)-2)))

  Of OTHER NEXT_DAY (LAST_DAY (ADD_MONTHS (TRUNC(startdate,'MM'), ((LEVEL*2)-2))), "LUN")-7

  END AS lastmonday

  FROM (SELECT SYSDATE startdate,

  SYSDATE + 300 enddate

  THE DOUBLE)

  CONNECT BY LEVEL<=>

• use to insert the last day of each month

  Hi, I wrote this code but it did not insert the 20 December and it is January 1, and next time running is 31 January... I would like to work to make this insert every last day of each month...
  

  begin
    sys.dbms_job.submit(job => :job,
                        what => 'INSERT INTO hisrtu 
                                          SELECT rtunum, linked, dbchkd, chaqty, dbvald, ctrltr, meaadj, qtyrtu, month, year, entity, rtunam 
                                          FROM qtyrtu;',
                        next_date => to_date('31-10-2010 01:00:03', 'dd-mm-yyyy hh24:mi:ss'),
                        interval => 'SYSDATE +30');
    commit;
  end;
  /

  I don't know if it's because I assigned the date 31-10... In November it's just but December has not...
  
  Best regards, hope you can help me.
  
  I use oracle 9i

  Hello

  If you want the job to run at 02:00 the last day of the next month, then the desired argument is:

  interval => 'ADD_MONTHS (TRUNC (SYSDATE, ''MONTH''), 2) - (22 / 24)'
  

  In other words, if she finishes running at 02:30 on 31 January, then
  TRUNC (SYSDATE, 'MONTH') will be at 00:00:00 January 1
  ADD_MONTHS (TRUNC (SYSDATE, 'MONTH'), 1) will be March 1st at 00:00:00
  ADD_MONTHS (TRUNC (SYSDATE, 'MONTH'), - 1) (22 / 24) will be 22 hours earlier, i.e. 02:00 on the last day of February.

  logandro wrote:
  I don't know if it's because I assigned the date 31-10... In November it's just but December has not...

  Do you mean this line?

  next_date-online to_date (October 31, 2010 01:00:03 ',' dd-mm-yyyy hh24:mi:ss'),

  I don't think that's a problem. The next_date specifies only when it work baptisms. I guess he ran on October 31; you said it was running well on 30 November, 30 days after that. Something has to happen after November 30.

  Note that the interval argument is relative when the task completes. If a task takes 30 minutes to run, then

  interval => 'SYSDATE +30'
  

  means that, if it turns to
  01:00:03 October 31, then it will run again about
  01:30:03 November 30, around
  02:00:03 on December 30.
  02:30:03 January 29 and so on. If your business does that for a few seconds, it won't matter much.

  If you decide to use LAST_DAY, don't forget not that the hours-minutes-seconds referring LAST_DAY are the same as those of his argument.

• How to calculate the first day of the month following in BI Publisher

  I need to be able to calculate the first day of the month of a date field in a BI Publisher model. The field name is COL_CLM_180 and contains a date in the format "YYYY-MM-DD". I have found so far any documentation that still gives me an idea of what to do.

  Any help or advice would be appreciated

  Hi - thank you very much for your help. Your last answer got me going in the right direction and I indeed was dense. I don't know why I don't think to the calculation in the original query. I put the statement: TRUNC (ADD_MONTHS(A.COL_CLM_180,1), 'MM') in a column in the query of conduct and it has worked flawlessly. Thank you - thank you - thank you.

• number of days in the month

  Dear members,
  
  I have available dates,
  Let's say date1 = January 1, 2011"
  and date2 = April 15, 2011"
  
  My goal is to calculate the sum of the no. of days between this range.
  
  i.e.
  
  month = 01 days = 22
  month = 02 days-28
  month = 03 days = 31
  month = 04 days = 15
  
  so total days = 22 + 28 + 31 + 15 = 96 between individuals because of the date range.
  My goal is to have the figure 96 days that is the sum of the days of the month.
  
  
  
  Kind regards
  
  
  
  Teefu
  Developer
  
  Published by: teefu on July 21, 2011 10:09

  Hello

  If all you want is the total number of days, then you can simply subtract the earlier DATE of the last one, like this:

  SELECT     TO_DATE ( '15-04-2011'
            , 'DD-MM=YYYY'
            ) + 1 - TO_DATE ( '10-01-2011'     -- Jan. 10; '01-10-' would be Oct. 1
                      , 'DD-MM-YYYY'
                      )     AS days_diff
  FROM     dual;
  

  If you want to see a breakdown by month, here's one way:

  WITH     parameters     AS
  (
       SELECT     TO_DATE ( '10-01-2011'
                 , 'DD-MM-YYYY'
                 )     AS start_dt
       ,     TO_DATE ( '15-04-2011'
                 , 'DD-MM-YYYY'
                 )     AS end_dt
       FROM     dual
  )
  SELECT       TO_CHAR ( start_dt + LEVEL - 1
              , 'FMMonth YYYY'
              )     AS month
  ,       COUNT (*)     AS days
  FROM       parameters
  CONNECT BY     LEVEL <= 1 + end_dt - start_dt
  GROUP BY  TO_CHAR ( start_dt + LEVEL - 1
              , 'FMMonth YYYY'
              )
  ORDER BY  MIN (LEVEL)
  ;
  

  Output:

  MONTH                      DAYS
  -------------------- ----------
  January 2011                 22
  February 2011                28
  March 2011                   31
  April 2011                   15
  

  Published by: Frank Kulash, July 21, 2011 01:10

• First day of the month between 08:00 and 18:30

  DB version: 10.2.0.4
  
  Due to the nature of our the first working day of each month (like today 1 MS), the DB will be very busy.
  We execute a proc purge planned Windows Scheduler every 5 minutes every day.
  
  But the first day of every month (like today 1 MS), if it's a day of the week (Monday to Friday) we do not want this proc running between 08:00 and for dinner from 18:30.
  
  So, I thought including the logic in the procedure itself.
  
  How can I implement this?
  
  Edited by: user872043, Sep 1 2010 08:48

  Here is an example of using DBMS_SCHEDULER which should meet your needs:

  Create schedules

  BEGIN
          DBMS_SCHEDULER.CREATE_SCHEDULE
          ( schedule_name   => 'every_5_schedule'
          , repeat_interval => 'FREQ=MINUTELY; BYMONTHDAY=2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31; INTERVAL=5; BYSECOND=0'
          , comments        => 'Schedule to run every 5 minutes'
          );
  
          DBMS_SCHEDULER.CREATE_SCHEDULE
          ( schedule_name   => 'busy_hourly_schedule'
          , repeat_interval => 'FREQ=MINUTELY; BYMONTHDAY=1; BYHOUR=0,1,2,3,4,5,6,7,19,20,21,22,23; INTERVAL=5;'
          , comments        => 'Schedule to run every 5 minutes except certain hours on the first of the month'
          );
  
          DBMS_SCHEDULER.CREATE_SCHEDULE
          ( schedule_name   => 'busy_6pm_schedule'
          , repeat_interval => 'FREQ=MINUTELY; BYMONTHDAY=1; BYHOUR=18; BYMINUTE=30,35,40,45,50,55'
          , comments        => 'Schedule to run every 5 minutes from 6:30 - 7'
          );
  END;
  /
  

  There are three lists.

  every_5_schedule - runs every 5 minutes for all the days that are not on the first day of the month
  busy_hourly_schedule - runs every 5 minutes for the acceptable hours on the first day of the month
  busy_6pm_schedule - retrieves the half hour from 6:30 - 7 on the first day of the month.

  Then, create your business:

  Job creation

  BEGIN
          DBMS_SCHEDULER.CREATE_JOB
          ( job_name        => 'YOUR JOB'
          , job_type        => 'PLSQL_BLOCK'
          , job_action      =>'BEGIN YOURPROC; END;'
          , repeat_interval => 'EVERY_5_SCHEDULE, BUSY_HOURLY_SCHEDULE, BUSY_6PM_SCHEDULE'
          , enabled         => TRUE
          , comments        => 'Some comments'
          );
  END;
  /
   
  

  HTH!