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Top Forums Shell Programming and Scripting Date command is not working properly Post 302895250 by Scrutinizer on Sunday 30th of March 2014 06:34:00 AM
Old 03-30-2014
Hi, try:
Code:
date --date="$(date +%Y%m15) 1 month ago" +%Y%m

This behavior occurs, because februari 30 does not exist. So it is best to use the middle of the month to calculate the previous month..


--
Interestingly in recent ksh93 this seems to work alright:
Code:
$ printf "%(%Y%m)T\n" "last month"
201402


Last edited by Scrutinizer; 03-30-2014 at 11:40 AM..
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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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