问题
I am looking to get previous date in unix / shell script .
I am using the following code
date -d ’1 day ago’ +’%Y/%m/%d’
But I am getting the following error.
date: illegal option -- d
As far as I've read on the inetrnet , it basically means I am using a older version of GNU. Can anyone please help with this.
Further Info
unix> uname -a
SunOS Server 5.10 Generic_147440-19 sun4v sparc SUNW,Sun-Fire-T200
Also The below command gives an error.
unix> date --version
date: illegal option -- version
usage: date [-u] mmddHHMM[[cc]yy][.SS]
date [-u] [+format]
date -a [-]sss[.fff]
回答1:
Several solutions suggested here assume GNU coreutils being present on the system. The following should work on Solaris:
TZ=GMT+24 date +’%Y/%m/%d’
回答2:
try this:
date --date="yesterday" +%Y/%m/%d
回答3:
dtd="2015-06-19"
yesterday=$( date -d "${dtd} -1 days" +'%Y_%m_%d' )
echo $yesterday;
回答4:
you can use
date -d "30 days ago" +"%d/%m/%Y"
to get the date from 30 days ago, similarly you can replace 30 with x amount of days
回答5:
In order to get 1 day back date using date command:
date -v -1d It will give (current date -1) means 1 day before .
date -v +1d
This will give (current date +1) means 1 day after.
Similarly below written code can be used in place of d to find out year,month etc
y-Year,
m-Month
w-Week
d-Day
H-Hour
M-Minute
S-Second
回答6:
the following script prints the previous date to the targetDate (specified Date or given date)
targetDate=2014-06-30
count=1
startDate=$( echo `date -d "${targetDate} -${count} days" +"%Y-%m-%d"`)
echo $startDate
回答7:
SunOS ships with legacy BSD userland tools which often lack the expected modern options. See if you can get the XPG add-on (it's something like /usr/xpg4/bin/date) or install the GNU coreutils package if you can.
In the meantime, you might need to write your own simple date handling script. There are many examples on the net e.g. in Perl. E.g. this one:
vnix$ perl -MPOSIX=strftime -le 'print strftime("%Y%m", localtime(time-86400))'
201304
(Slightly adapted, if you compare to the one behind the link.)
回答8:
I have used the following workaround to get to the required solution .
timeA=$(date +%Y%m)
sysD=$(date +%d)
print "Initial sysD $sysD">>LogPrint.log
sysD=$((sysD-1))
print "Final sysD $sysD">>LogPrint.log
finalTime=$timeA$sysD
I hope this is useful for people who are facing the same issue as me.
回答9:
$ date '+%m/%d/%Y' --- current date
$ TZ=Etc/GMT+24 date '+%m/%d/%Y' -- one dayprevious date
Use time zone appropriately
回答10:
Try This: gdate -d "1 day ago" +"%Y/%m/%d"
回答11:
Problem
You are using backticks, rather than single quotes, for your arguments. You may also not be using GNU date, or a version of date that supports the flag you are using.
Solution
Quote your arguments properly. For example:
$ date -d '1 day ago' +'%Y/%m/%d'
2013/04/14
来源：https://stackoverflow.com/questions/16027776/getting-a-previous-date-in-bash-unix