How subtract days from date in bash shell on Solaris 11?

Question

I've been trying subtrac some days $days from a date $date with format yyyy-MM-dd, but nothing has worked on Solaris 11. Some solution is a 'trick' with the timezone, but it depends on the timezone and I think that is exactly that, a trick.

I would like a cheaper solution, because the only thing I can think is to convert the date to julian representation and then subtract one day and again obtain yyyy-MM-dd representation, for example:

date=2000-12-31
days=1
julian=$(toJulian $date)
resultJulian=$(subtractDays $julian $days)
resultGregorian=toGregorian $resultJulian

So, how can I do it without all this proccess? Thanks.

Answer 1

If you don't have GNU date or GNU awk, consider perl:

subtractDays() {
  local date numDays
  date=$1
  numDays=$2

  date=$date days=$numDays perl -e '
    use Env qw(date days);
    use Time::Piece;
    use Time::Seconds;

    my $start_time = Time::Piece->strptime($date, "%Y-%m-%d");
    my $end_time = $start_time - (ONE_DAY * $days);
    print $end_time->ymd . "\n";'
}

...thereafter:

subtractDays 2000-12-31 1

...emits...

2000-12-30