YYYY-MM-DD format date in shell script

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

Try: $(date +%F)

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

You can do something like this:

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

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)

$(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