On Sun, 31 Aug 2008 02:52:07 +0300 "Ivan \"Rambius\" Ivanov" <[EMAIL PROTECTED]> wrote:
> I need to format the current date (as returned by date(1) ) to the > pattern m-d-yyyy, where m is the month in one or digits, d is the day > in one or two digits, and yyyy is the year in four digits. The problem > for me is the day and the month, for example August should be 8, and > not 08, and 5th of September should be 9-5-2008 and not 09-05-2008. > hello rambius! you can give this script a try - it seems to do what you want and has comments too. save it as de0.sh, chmod +x it and run it as ./de0.sh `date "+%m-%d-%Y"` (there are no doubt better ways to do what you want especially if you use a more advanced shell like zsh, but this may be sufficient) ================== #!/bin/sh # removes 0 from mm-dd-yyyy # run with ./de0.sh `date "+%m-%d-%Y"` #the whole date from argument $1 mmddyyyy=$1 #get the year yyyy=${mmddyyyy##*-} #get the month and day mmdd=${mmddyyyy%-*} #get the day dd=${mmdd#*-} #get the month mm=${mmdd%-*} #remove 0 if only at beginning of month, day and add on the year echo ${mm#0}-${dd#0}-$yyyy ================== -- In friendship, prad ... with you on your journey Towards Freedom http://www.towardsfreedom.com (website) Information, Inspiration, Imagination - truly a site for soaring I's _______________________________________________ freebsd-questions@freebsd.org mailing list http://lists.freebsd.org/mailman/listinfo/freebsd-questions To unsubscribe, send any mail to "[EMAIL PROTECTED]"