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

In bash (>=4.2) it is preferable to use printf's built-in date formatter (part of bash) rather than the external date (usually GNU date).

As such:

# put current date as yyyy-mm-dd in $date
# -1 -> explicit current date, bash >=4.3 defaults to current time if not provided
# -2 -> start time for shell
printf -v date '%(%Y-%m-%d)T\n' -1 

# put current date as yyyy-mm-dd HH:MM:SS in $date
printf -v date '%(%Y-%m-%d %H:%M:%S)T\n' -1 

# to print directly remove -v flag, as such:
printf '%(%Y-%m-%d)T\n' -1
# -> current date printed to terminal

In bash (<4.2):

# put current date as yyyy-mm-dd in $date
date=$(date '+%Y-%m-%d')

# put current date as yyyy-mm-dd HH:MM:SS in $date
date=$(date '+%Y-%m-%d %H:%M:%S')

# print current date directly
echo $(date '+%Y-%m-%d')

Other available date formats can be viewed from the date man pages (for external non-bash specific command):

man date

Answer 2

Try: $(date +%F)

The %F option is an alias for %Y-%m-%d

Answer 3

You can do something like this:

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

Answer 4

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 5

$(date +%F)

output

2018-06-20

Or if you also want time:

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

can be used to remove colons (:) in between

output

2018-06-20_09-55-58

Answer 6

date -d '1 hour ago' '+%Y-%m-%d'

The output would be 2015-06-14.

Answer 7

With recent Bash (version ≥ 4.2), you can use the builtin printf with the format modifier %(strftime_format)T:

$ printf '%(%Y-%m-%d)T\n' -1  # Get YYYY-MM-DD (-1 stands for "current time")
2017-11-10
$ printf '%(%F)T\n' -1  # Synonym of the above
2017-11-10
$ printf -v date '%(%F)T' -1  # Capture as var $date

printf is much faster than date since it's a Bash builtin while date is an external command.

As well, printf -v date ... is faster than date=$(printf ...) since it doesn't require forking a subshell.

Answer 8

I use the following formulation:

TODAY=`date -I`
echo $TODAY

Checkout the man page for date, there is a number of other useful options:

man date

Answer 9

if you want the year in a two number format such as 17 rather than 2017, do the following:

DATE=`date +%d-%m-%y`

Answer 10

I use $(date +"%Y-%m-%d") or $(date +"%Y-%m-%d %T") with time and hours.

Answer 11

Whenever I have a task like this I end up falling back to

$ man strftime

to remind myself of all the possibilities for time formatting options.

Answer 12

Try to use this command :

date | cut -d " " -f2-4 | tr " " "-" 

The output would be like: 21-Feb-2021

Answer 13

#!/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 14

I used below method. Thanks for all methods/answers

ubuntu@apj:/tmp$ datevar=$(date +'%Y-%m-%d : %H-%M')
ubuntu@apj:/tmp$ echo $datevar
2022-03-31 : 10-48

Answer 15

Try this code for a simple human readable timestamp:

dt=$(date)
echo $dt

Output:

Tue May 3 08:48:47 IST 2022

Answer 16

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} -------------"