getting a previous date in bash/unix

StackOverflow https://stackoverflow.com/questions/16027776

Question

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]
Was it helpful?

Solution 5

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’

OTHER TIPS

try this:

date --date="yesterday" +%Y/%m/%d

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

dtd="2015-06-19"
yesterday=$( date -d "${dtd} -1 days" +'%Y_%m_%d' )
echo $yesterday;

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

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

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.)

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.

$ date '+%m/%d/%Y' --- current date


$ TZ=Etc/GMT+24 date  '+%m/%d/%Y'  -- one dayprevious date

Use time zone appropriately

Try This: gdate -d "1 day ago" +"%Y/%m/%d"

date -d "yesterday" '+%Y-%m-%d'

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
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