I have one file with below entry. There are multiple entries, but for sample I used just three lines.
my requirment is to create a script by which it will pull only those entries which modification time is greater than 2 weeks (or 15 days). if I run script today, it will compare date from today and pull entriies which modification time is greater than 2 weeks.
Is there any command or combination of commands which can be used for this activity?
Input
Required Output
Below is Output of date command in my server
Last edited by vbe; 08-17-2015 at 05:35 AM..
Reason: code tags please. not HTML...
I want to get previous date from date command. I am using ksh shell.
Exmp:
today is 2008.09.04
I want the result : 2008.09.03
Please help.
Thanks in advance. (4 Replies)
I did some searches, but couldn't really find what I'm looking for. I have a file formatted as below:
BOF ABC CO - XYZ COMM DATA OF 07/05/2011
EBA00000001 sdfa rtyus uyml
EBB00000001 54682 984w3
EBA00000002 mkiyuasdf 98234
I want to pull the date from the header record and add it... (4 Replies)
hey guys.
the following line is a line taken from apache's access_log
10.10.10.10 - jdoe "GET /images/down.gif HTTP/1.1" 304
I'm concerned about the field that has the date and time in it.
if assuming the delimiter in the file is a space, then the fourth field will always have the date... (5 Replies)
Hi
I want to get tomorrow and yesterday date from date command. My shell is KSH and server is AIX. I tried several options, but unable to do. Please help on this.
Regards
Rajesh (5 Replies)
Hi,
In order to make our debugging easier in log files, I need this script.
My log file will be structured like this :
------Invoking myfile -param:start_time=1371150900000 -param:end_time=1371151800000 for 06/14/2013
<multiple lines here>
.....
- Step Sybase CDR Table.0 ended... (3 Replies)
Hi all,
Need an urgent help on the below scenario.
script:
awk -F","
'BEGIN { #some variable assignment}
{ #some calculation and put values in array}
END {
year=#getting it from array and assume this will be 2014
month=#getting it from array and this will be 05
date=#... (7 Replies)
HI,
Can anyone tell me how to pull the date and file name separated by a space using the find command or any other command. I want to look through several directories and based on a date timeframe (find -mtime -7), output the file name (without the path) and the date(in format mmddyyyy) to a... (2 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)
Discussion started by: kmarcus
20 Replies
LEARN ABOUT OSX
time::seconds
Time::Seconds(3pm) Perl Programmers Reference Guide Time::Seconds(3pm)NAME
Time::Seconds - a simple API to convert seconds to other date values
SYNOPSIS
use Time::Piece;
use Time::Seconds;
my $t = localtime;
$t += ONE_DAY;
my $t2 = localtime;
my $s = $t - $t2;
print "Difference is: ", $s->days, "
";
DESCRIPTION
This module is part of the Time::Piece distribution. It allows the user to find out the number of minutes, hours, days, weeks or years in a
given number of seconds. It is returned by Time::Piece when you delta two Time::Piece objects.
Time::Seconds also exports the following constants:
ONE_DAY
ONE_WEEK
ONE_HOUR
ONE_MINUTE
ONE_MONTH
ONE_YEAR
ONE_FINANCIAL_MONTH
LEAP_YEAR
NON_LEAP_YEAR
Since perl does not (yet?) support constant objects, these constants are in seconds only, so you cannot, for example, do this: "print
ONE_WEEK->minutes;"
METHODS
The following methods are available:
my $val = Time::Seconds->new(SECONDS)
$val->seconds;
$val->minutes;
$val->hours;
$val->days;
$val->weeks;
$val->months;
$val->financial_months; # 30 days
$val->years;
$val->pretty; # gives English representation of the delta
The usual arithmetic (+,-,+=,-=) is also available on the objects.
The methods make the assumption that there are 24 hours in a day, 7 days in a week, 365.24225 days in a year and 12 months in a year.
(from The Calendar FAQ at http://www.tondering.dk/claus/calendar.html)
AUTHOR
Matt Sergeant, matt@sergeant.org
Tobias Brox, tobiasb@tobiasb.funcom.com
BalieXXzs SzabieXX (dLux), dlux@kapu.hu
LICENSE
Please see Time::Piece for the license.
Bugs
Currently the methods aren't as efficient as they could be, for reasons of clarity. This is probably a bad idea.
perl v5.16.2 2012-10-11 Time::Seconds(3pm)