How to find the date of previous day in shell script?


 
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Old 07-16-2013
How to find the date of previous day in shell script?

Hi Experts,

i am using the below code get the date of previous day.

#!/usr/bin/ksh

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

When i execute the code i am getting output as:

20130715

What i am trying here is, based on the date passed i am fetching previus day's date.

for example:

If date passed is date = 16/07/2013, i need date of previous day

15/07/2013 but now i am getting output as 20130715, how can i convert this to format 15/07/2013

Thanks in advance.

Last edited by Scott; 07-16-2013 at 03:52 AM.. Reason: Double post. Closed.
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CAL_FROM_JD(3)								 1							    CAL_FROM_JD(3)

cal_from_jd - Converts from Julian Day Count to a supported calendar

SYNOPSIS
array cal_from_jd (int $jd, int $calendar) DESCRIPTION
cal_from_jd(3) converts the Julian day given in $jd into a date of the specified $calendar. Supported $calendar values are CAL_GREGORIAN, CAL_JULIAN, CAL_JEWISH and CAL_FRENCH. PARAMETERS
o $jd - Julian day as integer o $calendar - Calendar to convert to RETURN VALUES
Returns an array containing calendar information like month, day, year, day of week, abbreviated and full names of weekday and month and the date in string form "month/day/year". EXAMPLES
Example #1 cal_from_jd(3) example <?php $today = unixtojd(mktime(0, 0, 0, 8, 16, 2003)); print_r(cal_from_jd($today, CAL_GREGORIAN)); ?> The above example will output: Array ( [date] => 8/16/2003 [month] => 8 [day] => 16 [year] => 2003 [dow] => 6 [abbrevdayname] => Sat [dayname] => Saturday [abbrevmonth] => Aug [monthname] => August ) SEE ALSO
cal_to_jd(3), jdtofrench(3), jdtogregorian(3), jdtojewish(3), jdtojulian(3), jdtounix(3). PHP Documentation Group CAL_FROM_JD(3)