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Top Forums Shell Programming and Scripting Substract date (month) Problem Post 302285995 by Whiteboard on Tuesday 10th of February 2009 05:23:09 AM
Old 02-10-2009
Try this

Code:

day=`date +%d`
month=`date +%m`
year=`date +%Y`
lmonth=`expr $month - 1`
if test "$lmonth" = "0"
then
lmonth=12
year=`expr $year - 1`
fi
echo "Last month was $lmonth/$year"
 

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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)
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