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

Similar Questions

• 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;

• 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>
  

• 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

• 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;
  

• 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

• Function to retrieve the first day of the month.

  Select (sysdate, 'IW') of double
  
  I know it takes the date of Monday. Is there a function like above that has the date of return to the first Monday of the month?
  
  Thanks in advance.

  Hello

  NEXT_DAY is very convenient, but it is dependent on the NLS_DATE_LANGUAGE, and you cannot pass an argument to override that.
  TRUNC (dt, 'IW') is not in the NLS settings, then you may prefer this:

  TRUNC ( 6 + TRUNC (SYSDATE, 'MONTH')
        , 'IW'
        )          AS first_monday
  

  1st Monday of the month is fixed to the Monday or before the 7th of the month.

• Monday the 1st business day of the month of 15 before and after 15.

  I want to choose the 1st Monday with the business day (Y) of a month (january), owned by the 15th of the month and after the 15th of the month.
  Whereas it should be 2 folders, 1 record should be before the 15th of the month with the word 'Monday' & Holiday (N) and 2nd record should be based on the 15th of the month
  with the word 'Monday' & Holiday (N).
  
  Table:-holiday
  Column
  ---------
  Time_Date - holiday - Time_Day
  2011-01-01 00:00:00 - N - Tuesday
  
  Please let me know the oracle query.
  
  Thanks and greetings
  R Rodriguez
  
  Edited by: 880353 August 19, 2011 05:14

  It's almost like this... just a matter of playing with the SQL to reorganize the command:

  Select
  min (case when TIME_DATE between trunc(sysdate,'MM') and trunc (sysdate, 'MM') + 13 and)
  H.HOLIDAY = ' only then end TIME_DATE to another null) «FIRST»,
  min (case when TIME_DATE between trunc (sysdate, 'MM') + 14 and last_day (sysdate) and)
  H.HOLIDAY = ' only then end TIME_DATE to another null) "FIRST > = 15.
  Holiday h.
  (select next_day (trunc(sysdate, 'MONTH')-1, 'Monday') + 7 *(level-1) double DD
  connect by level<= 5)="">
  where M.DD = h.TIME_DATE

• What day of the month are updates released and how often?

  I saw a calendar, but I struggle to understand. I am putting together a Calendar Update for distribution - to my company and the need to update dates scheduled for Firefox ESR 31.x. thank you Jeff

  Hi jeff, regular updates will be released every 6 weeks. you could use or subscribe to this public google calendar:
  https://www.Google.com/calendar/embed?src=Mozilla.com_2d37383433353432352d3939%40resource.Calendar.Google.com

• Every first of the month between Date1 and Date2

  I created this query... Basically, it selects the dates more low rate and high and every day during this period, he made a subquery by selecting incoming and outgoing so far.
  
  I need to make a selection by month, not per day. How can I change the temporary table of DAYS so that each first day of the MONTH is selected instead of DAILY
  
  
  I have this
  ---------------------
  with
  DATES in the form
  (select (select min (REC_DATE) of the STOCK) as FROM_DT, (select max (CHANGE_DATE) of STOCK_ADJUST) as DOUBLE TO_DT),
  DAYS like
  (select Connect FROM_DT + ((ROWNUM-1)) as DATES by ROWNUM DT < = ((TO_DT-FROM_DT)) + 1)
  
  Select
  DAYS. GTO
  (select sum (stm.qty_rec) of stock stm WHERE pnm_auto_key = 1 and STM. REC_DATE < DAYS. Pnm_auto_key GROUP BY DT) ENTERING,
  (select sum (saj.qty_adj) in stock_adjust saj
  join in-house stock stm on saj.stm_auto_key = stm.stm_auto_key
  inner join parts_master MFN on stm.pnm_auto_key = pnm.pnm_auto_key where pnm_auto_key = 1 and SAJ. CHANGE_DATE < DAYS. Group by pnm_auto_key DT) OUTGOING
  of DAYS;
  ---------------------
  
  I need it
  ---------------------
  with
  DATES in the form
  (select (select min (REC_DATE) of the STOCK) as FROM_DT, (select max (CHANGE_DATE) of STOCK_ADJUST) as DOUBLE TO_DT),
  MONTHS as
  xxx?
  
  Select
  MONTHS. GTO
  (select sum (stm.qty_rec) of stock stm WHERE pnm_auto_key = 1 and STM. REC_DATE < MONTH. Pnm_auto_key GROUP BY DT) ENTERING,
  (select sum (saj.qty_adj) in stock_adjust saj
  join in-house stock stm on saj.stm_auto_key = stm.stm_auto_key
  inner join parts_master MFN on stm.pnm_auto_key = pnm.pnm_auto_key where pnm_auto_key = 1 and SAJ. CHANGE_DATE < MONTH. Group by pnm_auto_key DT) OUTGOING
  of MONTHS;
  ---------------------

  I won't have to try to understand which shows your code not formatted, but maybe this will give you assistance...

  SQL> ed
  Wrote file afiedt.buf
  
    1  with t as (select to_date('05/01/2008','DD/MM/YYYY') as start_dt,
    2                    to_date('05/11/2008','DD/MM/YYYY') as end_dt from dual)
    3  --
    4  select add_months(trunc(start_dt,'MM'),rownum-1) as dt
    5  from t
    6* connect by rownum <= months_between(end_dt, start_dt)+1
  SQL> /
  
  DT
  ---------
  01-JAN-08
  01-FEB-08
  01-MAR-08
  01-APR-08
  01-MAY-08
  01-JUN-08
  01-JUL-08
  01-AUG-08
  01-SEP-08
  01-OCT-08
  01-NOV-08
  
  11 rows selected.
  
  SQL>
  

  Published by: BluShadow on November 13, 2008 12:52

• 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.

• 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
  

• 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