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