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I can take a date and convert is to an integer.
date_as_integer = $("2020-06-13" | sed "s/-//g")
However is there a way to convert it back to date?
Answer
Doing it with string manipulation using bash parameter expansion:
date2int() {
printf "%sn" "${1//-/}"
}
int2date() {
printf "%s-%s-%sn" "${1:0:4}" "${1:4:2}" "${1:6:2}"
}
date +%F # => 2020-07-24
n=$(date2int "$(date +%F)"); echo "$n" # => 20200724
d=$(int2date "$n"); echo "$d" # => 2020-07-24
This technique is definitely NOT OK for date arithmetic.
# subtract 2 days from Jan 1 date="2000-01-01" n=$(date2int "$date") int2date "$((n - 2))" # => 2000-00-99
You might as well stick to GNU date
date="2000-01-01" date -d "$date - 2 day" "+%F" # => 1999-12-30
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