Bash script/command to print out date 5 min before/after

Question

I need to somehow use the date command in bash or another utility to print out the date and time, 5 minutes before and 5 minutes after a given value. For example:

input:

Thu Dec 19 14:10

output:

Thu Dec 19 14:05 Thu Dec 19 14:10 Thu Dec 19 14:15

I see that the date command can be used to do this on the current date/time, can it be used with a passed value, i.e. not the current date/time?

Answer 1

You can achieve this, for the current time, by typing.

$ date --date='5 minutes ago'; date; date --date='5 minutes'
Qui Dez 19 16:09:17 BRST 2013
Qui Dez 19 16:14:17 BRST 2013
Qui Dez 19 16:19:17 BRST 2013

To use a specific date (ex 1978/01/10).

$ date --date='1978-01-10 + 5 minutes'
Ter Jan 10 00:05:00 BRT 1978

Answer 2

With GNU date, you can do a simple form of date/time arithmetic with the argument to the --date option:

$ date --date 'now + 5 minutes'

With BSD date (at least, the version that ships with Mac OS X), the -v option allows you to do something similar:

$ date -v +5M
$ date -v -5M