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). Note that invoking a subshell has performance problems in Cygwin due to a slow fork() call on Windows.

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