cron job first Sunday of each month
HelloI have a job to run the 1st Sunday of each month
33 7 1,73,4,5,6,7 * 0
but his run every Sunday and every 1st to the 7th of each month
a little confused, I thought that all checks must return true for cron job
Hello
Please see this:
http://www.linuxquestions.org/questions/linux-software-2/scheduling-a-cron-job-to-run-on-the-first-sunday-of-the-month-524720/
HTH
KK
Tags: Database
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How to get the date for the first Monday of each month
Dear members,
How to get the date for the first Monday of each month.
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Here is the list of the first Monday of the month of this year:
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Make the scenario for each interface and expand when the script option is planning.
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994122 wrote:
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p_date IN DATE
p_first_sunday DATE
p_second_saturday DATE
)
IS
BEGIN
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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
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(2) http://www.obinotes.com/2010/02/tip-to-get-firstday-lastday.html
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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...
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By day, I mean working Mon - Fri.
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Sorry, I initially responded to the question "what is the * first * work of the month ', which is a little easier.
Here's a way you can code in SQL:
ALTER SESSION SET NLS_DATE_FORMAT = 'DD-Mon-YYYY Dy'; WITH got_dates AS ( SELECT TRUNC ( ADD_MONTHS (SYSDATE, LEVEL) , 'MONTH' ) AS a_date FROM dual CONNECT BY LEVEL <= 12 ) -- The part above just generates some sample dates -- The part below is the actual query: -- SELECT a_date , CASE TO_CHAR ( a_date + 2 , 'fmDay' , 'NLS_DATE_LANGUAGE=ENGLISH' -- If needed ) WHEN 'Monday' THEN 4 WHEN 'Tuesday' THEN 3 WHEN 'Saturday' THEN 4 WHEN 'Sunday' THEN 4 ELSE 2 END + a_date AS third_workday FROM got_dates ORDER BY a_date ;
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Published by: Frank Kulash, October 4, 2012 17:56
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HTH!
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