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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]
888
To get yesterday's date in bash shell you can use the Linux GNU date version with date -d yesterday or the Unix/macOS date version using date -j -v -1d .
gives the exit status of the last command executed. After a script terminates, a $? from the command-line gives the exit status of the script, that is, the last command executed in the script, which is, by convention, 0 on success or an integer in the range 1 - 255 on error. #!/bin/bash echo hello echo $?
try this:
date --date="yesterday" +%Y/%m/%d
you can use
date -d "30 days ago" +"%d/%m/%Y"
to get the date from 30 days ago, similarly you can replace 30 with x amount of days
29
dtd="2015-06-19"
yesterday=$( 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
5
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