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

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

Answer 2

In Bash:

get year-month-day from date

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

get year-month-day hour:minute:second from date

DATE=`date '+%Y-%m-%d %H:%M:%S'`

Other available date formats can be viewed from the date man pages:

man date

Answer 3

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

The output would be 2015-06-14.

Answer 4

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 5

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

$ man strftime

to remind myself of all the possibilities.

Answer 6

Try: $(date +%F)