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

Similar Questions

• 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.

• get the first and last day of the year

  and razm.date > = to_date ("' 1.01 ' |: $VDATE |", ' dd.mm.yyyy ") AND razm.date < = to_date ('31.12.2010. ' || (": $VDATE |", ' dd.mm.yyyy ");
  
  but I get:
  ORA-01830: date format picture ends before converting all of the input string
  01830 00000 - "date format picture ends before converting all of the input string.
  * Cause:
  * Action:
  
  
  : $VDATE is entered as 2010 param. I need to get data from range 1.1.2001 to 31.12.2010
  
  THX

  Why don't you use:

  to_char(razm.date,'YYYY') = :VDATE
  

• 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;

• Last day of the year in fast formula

  Hi all
  
  I have to deal with something special, based on the last day of the current year.
  
  How can I get the last day of the year? Is there a function available in the fast formulas?
  
  Appreciate all entries...
  
  Thank you
  SRS

  What is

  :
  L_last_day = TO_DATE (' 31 - DEC-'+ TO_CHAR (PAY_PROC_PERIOD_END_DATE, 'YYYY'), 'DD-MON-YYYY')
  :

  Clive

• How to get the last day of the week?

  HII
  
  I can get the week number of calendar for a given date using
  
  SELECT to_char (to_date('04/04/2011','MM/DD/YYYY'), 'WW') FROM dual
  
  can any body tell me, how to get the last day of the week?
  
  and the answer should be: 04/08/2011(8th april)
  Thank you
  San
  
  Published by: sandeep9 on April 4, 2011 03:50

  Perhaps this...

  SQL> select trunc(sysdate,'WW')+6 from dual;
  
  TRUNC(SYSDATE,'WW')+
  --------------------
  08-APR-2011 00:00:00
  
  SQL>
  

• How to get the last day of the year

  Hi all
  
  Thanks in advance...
  
  How do I get the last day of the year that I'm passing date at run time.
  
  I can manage to get the first day of the year by
  SELECT TRUNC(SYSDATE,'YEAR') AS FDAY_YEAR
  of the double
  
  Thanks in advance
  
  Concerning
  Sachin

    1*  select ADD_MONTHS(trunc(sysdate,'yyyy'),12)-1 dd from dual
  SQL> /
  
  DD
  -----------
  31-DEC-2010
  

• How do the 3rd week (business day) of the month in a date range?

  Hello

  
  

  I'm trying to find the 3rd day of the work week of each month in a given range of dates.  How is it possible in a SQL query? Thank you in advance.

  Hello

  Here's one way:

  SELECT first_day + BOX

  WHEN TO_CHAR (first_day, "DY") IN ("play", "Fri", "SAT")

  THEN 4

  WHEN TO_CHAR (first_day, "DY") ("SUN")

  THEN 3

  2 ELSE

  END AS workday_3

  DE)

  SELECT ADD_MONTHS (first_month

  , LEVEL - 1

  ) AS first_day

  DE)

  SELECT TO_DATE ('Sep 2013', 'My YYYY') AS first_month

  , TO_DATE ('Jan 2014', 'Mon YYYY') AS last_month

  OF the double

  )

  CONNECT BY LEVEL<= 1="" +="" months_between="" (last_month,="">

  )

  ORDER BY workday_3

  ;

  In another thread,.

  https://community.Oracle.com/thread/3513808

  you said that you had problems using a WITH clause, so I used online views instead of CLAUSES.

  With the above parameters, the result is:

  WORKDAY_3

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

  Wed 04-Sep-2013

  Set of 3 October 2013

  November 5, 2013 Mar

  Wednesday, December 4, 2013

  Fri January 3, 2014

• How to display current date time of day previous same time and same with last day of the week with 24 hours in relation to the graphic line obiee 11g

  Hello
  

  
  

  Can someone help me with the problem I am facing to... my scenario is currently that we show reports from 00 hours up to 24 hours. But now the client wants to exact more than 24 hours for chart reports.

  This means that if the time is now 16:05 hours, then graphical report must display data for 16:00 Hrs from yesterday until 16:00 hours that day. Is the same for the last week day 24 hours as well. Means they need two lines in the chart. Please help me...

  Thank you

  Srini VIEREN Terry.Williams dileraco

  Hello

  Now using these variables filter your column of measure...

  

  Kind regards

  Naga

• Find first &amp; last day of the week of the given date

  Hi all

  I have the input query, below

  WITH TEM AS (SELECT 11 January 13 ' DT dual union SELECT June 16, 12 'from dual union SELECT 4 July 12' dual Union SELECT 9 January 13 'from dual union SELECT 10 January 13' dual Union SELECT 4 January 13 'from dual union SELECT 7 January 13' from dual union SELECT 4 June 13 'from dual union SELECT 8 January 13' double) TO_DATE (DT) SELECT D1 ,' WEEK ' | To_char (to_date (DT), 'IW-YYYY') tem WK;

  I would like to get output that is below start date week (Saturday) & the end of the week (Friday)

  DT	Period of the WEEK	Max Date in this week	Min Date this week
  June 16, 12	WEEK 24-2012	June 16, 12	22 June 12
  July 4, 12	WEEK 27-2012	30 June 12	July 6, 12
  January 4, 13	WEEK 01-2013	January 4, 13	29 December 12
  January 7, 13	WEEK 02-2013	5 January 13	January 11, 13
  January 8, 13	WEEK 02-2013	5 January 13	January 11, 13
  January 9, 13	WEEK 02-2013	5 January 13	January 11, 13
  January 10, 13	WEEK 02-2013	5 January 13	January 11, 13
  January 11, 13	WEEK 02-2013	5 January 13	January 11, 13
  June 4, 13	WEEK 23-2013	1 June 13	June 7, 13

  Indicate also if a function available in oracle for this one.

  Like this? I think in your required output, 3rd line data, MAX_DATE should be 29 December 12 and MIN_DATE should be January 4, 13.

  WITH TEM AS

  (SELECT 11 January 13 ' DT double Union)

  SELECT 16 June 12 ' Union double

  SELECT 4 July 12 ' Union double

  SELECT January 9, 13 ' Union double

  SELECT 10 January 13 ' Union double

  SELECT 4 January 13 ' Union double

  SELECT January 7, 13 ' Union double

  SELECT 4 June 13 ' Union double

  SELECT 8 January 13 ' double)

  SELECT D1 DT,

  WK WEEK_PERIOD,

  BOX WHEN (TRIM (TO_CHAR (D1, 'DAY')) = 'SATURDAY') THEN D1

  ANOTHER NEXT_DAY(D1,'SATURDAY')-7

  END AS MAX_DATE_IN_THIS_WEEK,

  BOX WHEN (TRIM (TO_CHAR (D1, 'DAY')) = 'FRIDAY') THEN D1

  OF OTHER NEXT_DAY(D1,'FRIDAY') END AS MIN_DATE_IN_THIS_WEEK

  DE)

  SELECT TO_DATE (DT) D1,' WEEK ' | To_char (to_date (DT), 'IW-YYYY') tem WK)

  ORDER BY DT;

  OUTPUT:

  DT WEEK_PERIOD MAX_DATE_ MIN_DATE_

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

  JUNE 16, 12 WEEK 24-2012 16 JUNE 12 22 JUNE 12

  4 JULY 12 WEEK 27-2012 JUNE 30, 12 6 JULY 12

  4 JANUARY 13 WEEK 01-2013 29 DECEMBER 12 JANUARY 4, 13

  JANUARY 7, 13 WEEK 02-2013, JANUARY 5, 13 JANUARY 11, 13

  JANUARY 8, 13 WEEK 02-2013, JANUARY 5, 13 JANUARY 11, 13

  JANUARY 9, 13 WEEK 02-2013, JANUARY 5, 13 JANUARY 11, 13

  10 JANUARY 13 WEEK 02-2013, JANUARY 5, 13 JANUARY 11, 13

  JANUARY 11, 13 WEEK 02-2013, JANUARY 5, 13 JANUARY 11, 13

  4 JUNE 13 WEEK 23-2013, JUNE 1, 13 JUNE 7, 13

  9 selected lines.

• get the last day of the week

  Hi friends,
  
  I have the below query that returns me the data as below:
  
  Select 'Label', to_number (to_char (to_date(start_date,'yyyy-mm-dd'), 'IW')) - to_number (to_char (add_months (sysdate,-6), 'IW')) + 1 as a weekly sum, (col1) val yyy_view
  where to_date(start_dt,'yyyy-mm-dd') between trunc (add_months(sysdate,-6)) and trunc (sysdate)
  Group of to_char (to_date(start_date,'yyyy-mm-dd'), 'IW')
  order of to_char (to_date(start_date,'yyyy-mm-dd'), 'IW')
  
  Label weekly val
  1 20 data
  2 24 data
  3 20 data
  4 28 data
  5 20 data
  6 24 data
  
  and so on...
  
  Now the requirement is to get the last date of the week for each week as below
  
  Label weekly val
  given 2011-08-30 20
  given 2011-08-23 24
  given 2011-08-16 20
  given 28 2011-08-09
  given 2011-08-02 20
  
  the weekly column should have the date of the last week. In the example above Wednesday is supposed to be the last date of the week.
  
  Kind regards
  Pradeep

  Missed the part "Wednesday is considered date of last week. The week is Thursday to Wednesday. If so:

  select  'Label',
           trunc(min(start_dt)) as day2,
          sum(col1 ) val
    from  yyy_view
    where start_dt between trunc(add_months(sysdate,-6)) and trunc(sysdate)
    group by trunc(start_date - 3,'IW')
    order by trunc(start_date - 3,'IW')
  /
  

  SY.

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  Then restarted and I tried to download the malware Microsoft Windows (KB890830) removal tool, tried twice and was redirected each time.

  Any suggestions?

  Hi JudiDue,

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  Reference: what is the browser hijacking?

  Hope the helps of information.

  Please post back and we do know.

   

• How to get the Date in month based on week of the month.

  Hi gurus,

  I have provided.

  Quarter (Q) = > 1 (Jan - Mar), 2 (April-June), 3 (July-September), 4 (October-December).

  Months (M) = > Jan = 1, Feb = 2, Mar = 3 in the 1st quarter.

  ARP = 1, may = 2, Jun = 3 in the 2nd quarter,

  Jul = 1, August = 2, Ms = 3 in Q3,

  Oct 1 Nov = = 2, Dec = 3 in the 4th quarter

  If I give Q = 3, M = 3, W = 1 day = Fri so I get 6-Sep-13

  Similarly, if I give Q = 2, M = 1, O = 3, day = game so I get 18 April 13.

  Could someone help me higher and higher.

  Concerning

  Sanjeev

  Believing that when you say you want to say that it is 1 - 5 in a month with the 1st day of the month as the beginning of the 1st week of the week. You can try this.

  SQL > t
  2 as
  (3)
  4. Select q 3, 3 m, 1 w, 'Ven' d
  5 double
  6 union
  7 all the
  8. Select q 2, 1 m, 3 w, d 'game '.
  9 double
  10)
  11. Select q, m, w, d, d_day
  12 years of)
  13 select y + (level - 1) d_day
  14, ceil (extract (day of cast (y + (level - 1) as timestamp)) / 7) week_7_day
  15                   , t.*
  16 of)
  17 select add_months (trunc (sysdate, 'year') (q * 3-3) +(m-1)) y
  18                             , t.*
  19 t
  20                     ) t
  21 connect
  22 by level<= last_day(y)="" -="" y="" +="">
  23 and prior y = y
  dbms_random.value (24 and prior) is not null
  25         )
  where the 26 w = week_7_day
  27 and substr (to_char (d_day 'day'), 1, 3) = lower (d);

  Q M W D D_DAY
  ---------- ---------- ---------- --- ---------
  2 1 3 Friday 18 April 13
  3 3 1 Fri 06-SEP-13

  SQL >

• by selecting specific days of the week for the date range?

  I have a table with a number of clients for about a year, but I want to only select Wednesday and Thursday of
  every week, since the beginning of the dates. Table is simple and has two columns. Each line is separate from a Date
  There is no duplicate of Date, it is counted for every day of the year.
  
  column 1: number of clients, count (customer_id)
  column 2: Date
  
  Need help with the best way to achieve this.
  
  Not sure if it is even possible to select a date in the name of the day?
  
  Basically, I want to select every Wednesday and Thursday of each week and compare the counts during the week, the week during
  week for the whole week see if charges go upwards or downwards, to get trends, thank you!

  Hello

  Kodiak_Seattle wrote:
  I have a table with a number of clients for about a year, but I want to only select Wednesday and Thursday of
  every week, since the beginning of the dates. Table is simple and has two columns. Each line is separate from a Date
  There is no duplicate of Date, it is counted for every day of the year.

  column 1: number of clients, count (customer_id)
  column 2: Date

  Need help with the best way to achieve this.

  Not sure if it is even possible to select a date in the name of the day?

  Sorry, we don't know what you want.

  To see if a date given (dt) is a Wednesday or Thursday, you can use:

  WHERE   TO_CHAR ( dt
            , 'DY'
            , 'NLS_DATE_LANGUAGE=ENGLISH'     -- If necessary
            )  IN ('WED', 'THU')
  

  I hope that answers your question.
  If not, post a small example data (CREATE TABLE and only relevant columns, INSERT statements) for all of the tables involved and also publish outcomes from these data.
  Explain, using specific examples, how you get these results from these data.
  Always say what version of Oracle you are using (for example, 11.2.0.2.0).
  See the FAQ forum {message identifier: = 9360002}

• How to calculate this column (the last week of the month)?

  I have a request where it shows a few months of calculations of sage
  
  ex. This report is for Dec-2010
  
  statement line. A reduced rate. Approved rate. Last week declined the rate |
  -------------------------
  Asia report | ------20%----- | -----45%-------- | --------4%------ |
  Africa report | ------44% -----| -----21%-------- | ------12%-------|
  
  I'm figuring the column rate declined last week . The logic is simply not coming to my mind.
  Can someone suggest me how can I calculate this?

  We'll see. This report at any time after the previous month running has closed...

  (1) TIMESTAMPADD (SQL_TSI_DAY, (DayOfMonth (CURRENT_DATE)) *-1, CURRENT_DATE) will give you the last day of the previous month.

  (2) TIMESTAMPADD (SQL_TSI_DAY, DAYOFWEEK (TIMESTAMPADD (SQL_TSI_DAY, (DayOfMonth (CURRENT_DATE)) *-1, CURRENT_DATE)) *-1, TIMESTAMPADD (SQL_TSI_DAY, (DayOfMonth (CURRENT_DATE)) *-1, CURRENT_DATE)) will give you the Saturday before the last week of the month.

  (3) TIMESTAMPADD (SQL_TSI_DAY, 1, TIMESTAMPADD (SQL_TSI_DAY, DAYOFWEEK (TIMESTAMPADD (SQL_TSI_DAY, (DayOfMonth (CURRENT_DATE)) *-1, CURRENT_DATE)) *-1, TIMESTAMPADD (SQL_TSI_DAY, (DayOfMonth (CURRENT_DATE)) *-1, CURRENT_DATE))) will give you the Sunday of the last week of the previous month.

  Now that you have the start of the last day of the previous week, you can create your filter... It will be "(between 3) and 1)" above... "