Date: Sat, 30 Aug 2008 18:47:10 -0700 From: prad <prad@towardsfreedom.com> To: freebsd-questions@freebsd.org Subject: Re: Formatting dates to a specific pattern Message-ID: <20080830184710.2a618fad@gom.home> In-Reply-To: <89ce7f740808301652g169fa4d2v3fe5eff06100e31e@mail.gmail.com> References: <89ce7f740808301652g169fa4d2v3fe5eff06100e31e@mail.gmail.com>
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On Sun, 31 Aug 2008 02:52:07 +0300
"Ivan \"Rambius\" Ivanov" <rambiusparkisanius@gmail.com> 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)
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