Hi All,
I have to kick off a script on every Monday to get some data from database for last week (Sunday thru Saturday).
If its Monday (12/12/2005)
Begin date will be Sunday - 12/4/2005
End date will be Saturday - 12/10/2005
The script might not kick off on some Mondays.
So my... (2 Replies)
Hi,
I have a requirement to define all Saturdays/Sundays of every month in a year as holidays. I am making use of Date::Manip package available in perl.
I tried writing a recurrence as :
0:1*3:7:0:0:0 --> this relation helps me to define 3rd Sunday of a year as sunday.
My requirement is... (1 Reply)
Hi all,
I've used various scripts in the past to work out the date last week from the current date, however I now have a need to work out the date 1 week from a given date.
So for example, if I have a date of the 23rd July 2010, I would like a script that can work out that one week back was... (4 Replies)
Hello gurus,
I am hoping someone can help me with the required code/script to make this work. I have the following file with records starting at line 4:
NETW~US60~000000000013220694~002~~IT~USD~2.24~20110201~99991231~01~01~20101104~... (4 Replies)
Hi All,
I have a requirement in my project where a batch runs on any day of a week. The input files will land to the server every Sunday. I need to read these files based on the current BUS_DAY (which falls any day of the week).
For e.g : if the BUS_DAY is 20120308, we need to derive the... (3 Replies)
Hi All,
Can you please let me know how to get the yesterday's date of the given date if the given date is Sunday?
I can't use GNU. I have the code to get the yesterday's date based on the system date.
Thanks (5 Replies)
current date command runs well
awk -v t="$(date +%Y-%m-%d)" -F "'" '$1 < t' myname.dat
subtract 30 days fails
awk -v t="$(date --date="-30days" +%Y-%m-%d)" -F "'" '$1 < t' myname.dat
awk command in hp unix subtract 30 days automatically from current date without date illegal option error... (20 Replies)
Hello All,
we what we call a parameter file (.txt) where my application read dynamic values when the job is triggered, one of such values are below:
abc.txt
------------------
line1
line2
line3
$$EDWS_DATE_INSERT=08-27-2019
line4
$$EDWS_PREV_DATE_INSERT=08-26-2019
I am trying to... (1 Reply)
Discussion started by: pradeepp
1 Replies
LEARN ABOUT CENTOS
cal
CAL(1) User Commands CAL(1)NAME
cal - display a calendar
SYNOPSIS
cal [options] [[[day] month] year]
DESCRIPTION
cal displays a simple calendar. If no arguments are specified, the current month is displayed.
OPTIONS -1, --one
Display single month output. (This is the default.)
-3, --three
Display prev/current/next month output.
-s, --sunday
Display Sunday as the first day of the week.
-m, --monday
Display Monday as the first day of the week.
-j, --julian
Display Julian dates (days one-based, numbered from January 1).
-y, --year
Display a calendar for the current year.
-V, --version
Display version information and exit.
-h, --help
Display help screen and exit.
PARAMETERS
A single parameter specifies the year (1 - 9999) to be displayed; note the year must be fully specified: cal 89 will not display a calendar
for 1989.
Two parameters denote the month (1 - 12) and year.
Three parameters denote the day (1-31), month and year, and the day will be highlighted if the calendar is displayed on a terminal. If no
parameters are specified, the current month's calendar is displayed.
A year starts on Jan 1. The first day of the week is determined by the locale.
The Gregorian Reformation is assumed to have occurred in 1752 on the 3rd of September. By this time, most countries had recognized the ref-
ormation (although a few did not recognize it until the early 1900's). Ten days following that date were eliminated by the reformation, so
the calendar for that month is a bit unusual.
HISTORY
A cal command appeared in Version 6 AT&T UNIX.
AVAILABILITY
The cal command is part of the util-linux package and is available from ftp://ftp.kernel.org/pub/linux/utils/util-linux/.
util-linux June 2011 CAL(1)