09-17-2008
Try:
Use date -d as shown below:
Quote:
date -d "$var days "
Wed Sep 24 00:30:27 IST 2008
var=60
date -d "$var days "
Mon Nov 17 00:30:37 IST 2008
10 More Discussions You Might Find Interesting
1. Shell Programming and Scripting
I have no idea why I can't get this to work, if anybody can help i would appreciate it.
#!/bin/bash
x=`cat counter.txt | wc -l`
y= '$x / 7'
printf "%d People have visited this page" $y
:confused: (2 Replies)
Discussion started by: paladyn_2002
2 Replies
2. Shell Programming and Scripting
I am using egrep to extract numbers from a file and storing them as variables in a script. But I am not able to do any arithmetic operations on the variables using "expr" because it stores them as char and not integers. Here is my code and the error I get. Any help will be appreciated.
#!/bin/sh... (3 Replies)
Discussion started by: emjayshaikh
3 Replies
3. UNIX for Dummies Questions & Answers
Hi, All,
I have a file, its content is as follows:
100 150
120 135
140 170
I want to insert a column, its content is determined by the difference between the two values in the same line, if the difference is less than 20, the new value is 1, otherwise is 0.
after the operation, the... (1 Reply)
Discussion started by: Jenny.palmy
1 Replies
4. Shell Programming and Scripting
I have output like following in a file
usmtnz-dinfsi19
72
71
38
1199
1199
0.8
19:23:58
usmtnz-dinfsi19
72
71
38
1199
1199
0.8
19:24:04 (9 Replies)
Discussion started by: fugitive
9 Replies
5. Shell Programming and Scripting
Hi
I have a file with 3 columns. say,
infile:
1 50 68
34 3 23
23 4 56
-------
-------
I want to generate n files from this file using a loop so that 1st column in output file is (column1 of infile/(2*n+2.561))
I am doing like this:
for ((i=1; i<=3; i++))
do
a=`echo... (3 Replies)
Discussion started by: Surabhi_so_mh
3 Replies
6. Shell Programming and Scripting
I am writing a script in zsh shell, it fetchs a number from a file using the awk command, store it as a variable, which in my case is a small number 0.62000. I want to change this number by multiplying it by 1000 to become 620.0 using the command in the script
var2=$((var1*1000))
trouble is... (2 Replies)
Discussion started by: piynik
2 Replies
7. UNIX for Dummies Questions & Answers
input:
Name|Operation
rec_10|1+2+2-
Output:
rec_10|1
Basically I am trying to calculate the result of "the path" in $3 where the operators follow the number and not preceding them like we normally do:
rec_10: +1+2-2=1
But I realise (I am sure there is a good reason for that) that awk... (7 Replies)
Discussion started by: beca123456
7 Replies
8. Shell Programming and Scripting
Hi,
Here is the script i try to perform arithmetic operation in two variables .
git branch -r | while read brname ; do
REV_COMMITS=`git rev-list --count $brname`
echo "$brname has $REV_COMMITS"
(( TOTAL = TOTAL + REV_COMMITS ))
echo "in loop" $TOTAL
done
echo "total is " $TOTAL
... (3 Replies)
Discussion started by: greet_sed
3 Replies
9. Shell Programming and Scripting
Hi,
I've this following text file
FileVersion = 1.03
Filetype = meteo_on_curvilinear_grid
TIME = 0 hours since 2016-10-03 12:00:00 +00:00
-6.855 -6.828 -6.801 -6.774 -6.747 -6.719 -6.691 -6.663 -6.634 -6.606 -6.577 -6.548 -6.519 -6.489
TIME = 0 hours since... (2 Replies)
Discussion started by: xisan
2 Replies
10. How to Post in the The UNIX and Linux Forums
I have to do some arithmetic operation on Field 8 which is calculated by Field 9/Field 7
Suppose i have data like :
0800123456|JAN|2017|JAN|2018|0800123456|0|0.0000|0.00|
0800234567|JAN|2017|JAN|2018|0800234567|4|2.5812|10.32|
0800666666|JAN|2017|JAN|2018|0800666666|2|1.7255|3.45|... (0 Replies)
Discussion started by: pumrao
0 Replies
LEARN ABOUT CENTOS
pam_lastlog
PAM_LASTLOG(8) Linux-PAM Manual PAM_LASTLOG(8)
NAME
pam_lastlog - PAM module to display date of last login and perform inactive account lock out
SYNOPSIS
pam_lastlog.so [debug] [silent] [never] [nodate] [nohost] [noterm] [nowtmp] [noupdate] [showfailed] [inactive=<days>]
DESCRIPTION
pam_lastlog is a PAM module to display a line of information about the last login of the user. In addition, the module maintains the
/var/log/lastlog file.
Some applications may perform this function themselves. In such cases, this module is not necessary.
If the module is called in the auth or account phase, the accounts that were not used recently enough will be disallowed to log in. The
check is not performed for the root account so the root is never locked out.
OPTIONS
debug
Print debug information.
silent
Don't inform the user about any previous login, just update the /var/log/lastlog file.
never
If the /var/log/lastlog file does not contain any old entries for the user, indicate that the user has never previously logged in with
a welcome message.
nodate
Don't display the date of the last login.
noterm
Don't display the terminal name on which the last login was attempted.
nohost
Don't indicate from which host the last login was attempted.
nowtmp
Don't update the wtmp entry.
noupdate
Don't update any file.
showfailed
Display number of failed login attempts and the date of the last failed attempt from btmp. The date is not displayed when nodate is
specified.
inactive=<days>
This option is specific for the auth or account phase. It specifies the number of days after the last login of the user when the user
will be locked out by the module. The default value is 90.
MODULE TYPES PROVIDED
The auth and account module type allows to lock out users which did not login recently enough. The session module type is provided for
displaying the information about the last login and/or updating the lastlog and wtmp files.
RETURN VALUES
PAM_SUCCESS
Everything was successful.
PAM_SERVICE_ERR
Internal service module error.
PAM_USER_UNKNOWN
User not known.
PAM_AUTH_ERR
User locked out in the auth or account phase due to inactivity.
PAM_IGNORE
There was an error during reading the lastlog file in the auth or account phase and thus inactivity of the user cannot be determined.
EXAMPLES
Add the following line to /etc/pam.d/login to display the last login time of an user:
session required pam_lastlog.so nowtmp
To reject the user if he did not login during the previous 50 days the following line can be used:
auth required pam_lastlog.so inactive=50
FILES
/var/log/lastlog
Lastlog logging file
SEE ALSO
pam.conf(5), pam.d(5), pam(8)
AUTHOR
pam_lastlog was written by Andrew G. Morgan <morgan@kernel.org>.
Inactive account lock out added by Toma Mraz <tm@t8m.info>.
Linux-PAM Manual 09/19/2013 PAM_LASTLOG(8)