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Top Forums Shell Programming and Scripting How to perform arithmetic operation on date Post 302237396 by dennis.jacob on Wednesday 17th of September 2008 03:02:02 PM
Old 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
 

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lchage(8)						      System Manager's Manual							 lchage(8)

NAME
lchage - Display or change user password policy SYNOPSIS
lchage [OPTION]... user DESCRIPTION
Displays or allows changing password policy of user. OPTIONS
-d, --date=days Set the date of last password change to days after Jan 1 1970. -E, --expire=days Set the account expiration date to days after Jan 1 1970. Set days to -1 to disable account expiration. -i, --interactive Ask all questions when connecting to the user database, even if default answers are set up in libuser configuration. -I, --inactive=days Disable the account after days after password expires (after the user user is required to change the password). -l, --list Only list current user's policy and make no changes. -m, --mindays=days Require at least days days between password changes. Set days to 0 to disable this checking. -M, --maxdays=days Require changing the password after days since last password change. Set days to 99999 to disable this checking. -W, --warndays=days Start warning the user days before password expires (before the user is required to change the password). EXIT STATUS
The exit status is 0 on success, nonzero on error. libuser Jan 12 2005 lchage(8)
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