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Help on Dates in Shell Scripting

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# 1  
Old 07-16-2013
Help on Dates in Shell Scripting
Hi Experts,

I am using the below code to get the previous day based on the date given as a input.

Code:
#!/usr/bin/ksh

datestamp=`date '+%Y%m%d'`
yest=$((datestamp -1))
echo $yest

Output: 20130714

How can i display the same output in 14/07/2013, i tired '+%d/%m/%y'` but i am getting error.

also how can i get the first day of the month based on the date which is fetched $yest.

Any help.

Thanks
Last edited by Scott; 07-16-2013 at 01:15 AM.. Reason: Code tags
# 2  
Old 07-16-2013
Do you have GNU date? just check man page of date if u have GNU date then you have option to get yesterday's date..

If not you have to modify TZ to get the desired date.. date - 1 wont work Smilie
# 3  
Old 07-16-2013
If your date command supports the -d option (gnu date)
you can do:

Yesterday:
Code:
date -d "-1 day" +%d/%m/%Y


1st of month:
Code:
YM=`date +%Y%m`
date -d ${YM}01 +%d/%m/%Y

# 4  
Old 07-16-2013
Hi Experts,

when i run the command in unix prompt: i got the below error:
Code:
date -d "-1 day" +%d/%m/%Y
date: illegal option -- d
Usage: date [-u] [+format]
       date [-u] [mmddhhmm[[cc]yy]]
       date [-a [-]sss.fff]

can you help me to sort this.

regards,
Last edited by Scott; 07-16-2013 at 01:15 AM.. Reason: Please use code tags
# 5  
Old 07-16-2013
OK you don't have gnu date you might have to use perl

yesterday.pl
Code:
#!/bin/perl
use POSIX; print strftime('%d/%m/%Y', localtime(time() - 24*60*60));

first.pl
Code:
#!/bin/perl
use POSIX;
my $year, $month;
($year, $month) = split(/\s/, strftime('%Y %m', localtime(time())));
printf strftime('%d/%m/%Y', localtime(mktime(0,0,0,1,$month-1,$year-1900,0,0)));

Last edited by Chubler_XL; 07-16-2013 at 01:39 AM..
# 6  
Old 07-16-2013
Hi Experts,

I want to use in unix script.

any help/suggestions

regards,
# 7  
Old 07-16-2013
For your first requirement, this should work
Code:
$ date -j -f "%s" $(($(date "+%s") - 86400))  "+%d/%m/%Y"
15/07/2013
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