YYYY-MM-DD format date in shell script

Question

I tried using $(date) in my bash shell script, however I want the date in YYYY-MM-DD format. How do I get this?

Answer 1

You can do something like this:

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

Answer 2

You're looking for ISO 8601 standard date format, so if you have GNU date (or any date command more modern than 1988) just do: $(date -I)

Answer 3

$(date +%F_%H-%M-%S)

can be used to remove colons (:) in between

output

2018-06-20_09-55-58

Answer 4

#!/bin/bash -e

x='2018-01-18 10:00:00'
a=$(date -d "$x")
b=$(date -d "$a 10 min" "+%Y-%m-%d %H:%M:%S")
c=$(date -d "$b 10 min" "+%Y-%m-%d %H:%M:%S")
#date -d "$a 30 min" "+%Y-%m-%d %H:%M:%S"

echo Entered Date is $x
echo Second Date is $b
echo Third Date is $c

Here x is sample date used & then example displays both formatting of data as well as getting dates 10 mins more then current date.

Answer 5

You can set date as environment variable and later u can use it

setenv DATE `date "+%Y-%m-%d"`
echo "----------- ${DATE} -------------"

or

DATE =`date "+%Y-%m-%d"`
echo "----------- ${DATE} -------------"