Day of the month Dimension or is there another way?

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• 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.

• First day of the month

  How to find the first day of the month if the month format is like TO_CHAR(SYSDATE,'YYYYMM')
  
  Sanjay

  Published by: user12957777 on May 23, 2012 21:12

  There is no 'day' in this date format.

  What you trying to do? What do you expect the result to look like?

  Are you wanting to set a date for the first day of the month and year of your example?

  SELECT TO_DATE(TO_CHAR(SYSDATE,'YYYYMM') || '01', 'YYYYMMDD') FROM DUAL;
  

• 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.
  
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  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!

• 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
  

• BEEP Scheduler - last day of the month

  Is there a a way to schedule reports to run on the last day of the month using the Scheduler of BEEP? I see that I can choose to run the monthly report and I choose the day to run it, but the last day of the month is different from one month to the other most of the time. Is it possible to do so in the Scheduler?
  
  Thank you!

  Oops, my bad,
  I did not review the request clearly.

  There is possibility to plan the report, because every month and he asks date of the month.
  If you register 31, his is not going to run it for each month, since used 31 are present in every month.

  Note: And there is no provision in the bi-editeur before you do :), front-end Bi Publisher accuses this support.
  You can use WebService at fire report every end of month :), but if you seek support front end 'calendar' homepage, start SR.

  As far as I know, its a bug, please contact technical support and connect a SR and let me know of the State.

• each first day of the month

  Hi friends here is the querry I wrote to display every first day of the month, but there is something wrong in it... Help me...
  create or replace procedure minmax is
  cursor minmax_cur is
  Select min (hiredate), max in emp (hiredate);
  number of v_minmonths;
  number of v_maxmonths;
  number of v_minyears;
  number of v_maxyears;
  x date of publication;
  date of y;
  
  Start
  Open minmax_cur;
  extraction minmax_cur
  x, y;
  v_minmonths: = extraction (x months);
  v_minyears: = extraction (year x);
  While (x, y) loop
  because me in 1... v_minmonths loop
  If (v_minmonths < 12) then
  v_minmonths: = v_minmonths + 1;
  x: = x + 28.
  dbms_output.put_line(v_minmonths ||) '/' || '01'. '/' ||
  v_minyears);
  on the other
  v_minmonths: = 0;
  v_minyears: = v_minyears + 1;
  x: = x + 28.
  
  end if;
  end loop;
  end loop;
  minMax end;
  and the results are
  
  01/01/1981
  01/02/1981
  01/03/1981
  ..................
  .........---
  01/04/1987
  01/05/1987
  01/06/1987
  01/07/1987
  01/08/1987
  01/09/1987
  01/10/1987
  01/11/1987 the problem is max (hiredate) is 5/23/1987...it must show up to 01/05/1987...
  I get a few more months...
  

• find the first day of the month

  Hello...
  can u pls let me know how to find the first day of the month, if the number of months (for ex - OCT-10) is passed as input
  
  Kind regards

  I'm not sure that there is a function called FIRST_DAY().

  SQL> SELECT first_day (SYSDATE)
    2    FROM DUAL;
  SELECT first_day (SYSDATE)
         *
  ERROR at line 1:
  ORA-00904: "FIRST_DAY": invalid identifier
  
  SQL> SELECT TRUNC (SYSDATE, 'month')
    2    FROM DUAL;
  
  TRUNC(SYS
  ---------
  01-DEC-08
  
  SQL> 
  

  I hope this helps.

  Kind regards
  Joice

• 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.

• query for the first day of the month

  Hi all

  How to get the first day of the month in the sql query.

  SQL> select trunc(sysdate,'mm') first_day from dual;
  
  FIRST_DAY
  ---------
  01-AUG-15
  
  SQL> select trunc(sysdate,'month') first_day  from dual;
  
  FIRST_DAY
  ---------
  01-AUG-15
  
  SQL> select trunc(sysdate,'mon')  first_day from dual;
  
  FIRST_DAY
  ---------
  01-AUG-15
  
  SQL>
  

• How to find the first day of the month following two years one year from now?

  Hello

  I need to find the first day of the month following two years ago, based on the current date.

  For example: If today is the 31/08/2015 so I need to find the date range between 31/08/2015 and on 09/01/2013. How can I write an SQL which allows to automate this calculation instead of hard-coding the values. Please do help me out with this so that I can build the SQL that can produce the desired result.

  Thank you

  Dhilip

  find the start of the month - 2 years in advance, early - 2 years + 1 months in advance and go back 1

  Select trunc (add_months(sysdate,24), 'MM'), trunc (add_months(sysdate,25), 'MM') - 1 double;

• first day of the month and time

  Hello

  Using oracle 11.2.0.3

  A table with a column date

  sample opening time data 04/08/2014-08:30 one single column, another column start date 04/02/2014

  Want to have a sql that will go through an opening time table set on the first day of the month as well as existing time

  for example expect times to be

  04/01/2014 08:30

  Code below will give the first day of the month

  What is the best way to achieve this-date data type columns.

  {code}

  Select trunc(start_time,'MM))

  of customer_hours ch

  {code}

  Select

  SYSDATE - trunc (sysdate) + trunc (sysdate, 'MM')

  of the double

  resp.

  Select

  start_time - trunc (start_time) + trunc (start_time, 'MM')

  of the double