mjhjmtu

I need to get the date and time of the command date minus 5 mins.

date:
29 Aug 2018 21:56:01

result:
29 Aug 2018 21:51:02

I tried to search on the Internet/forums but most of it involved using the -d option and what I have doesn't support the -d option.

This one works on Bash but I need something that works on ksh.

$ printf "%(%d %h %Y %H:%M:%S)Tn" $(( $(printf "%(%s)T") - 5 * 60 ))

You could use perl:

echo $(( $(printf "%(%s)T") - 5 * 60 )) | perl -lne 'print scalar localtime $_'

Or in pure perl:

perl -le 'print scalar localtime(time()-5*60)'

To format use strftime():

perl -MPOSIX=strftime -le 'print strftime("%d %h %Y %H:%M:%S", localtime(time()-5*60))'

bash actually copied the %(...)T format from ksh93 but only a small subset.

That's much more powerful in ksh93.

$ date; printf '%(%c)Tn' 'exactly five minutes ago'
Mon 3 Sep 08:08:32 BST 2018
Mon Sep 3 08:03:32 2018

bash only supports Unix epoch times (or -1 for now, though with recent versions you can also get now with a missing argument) while ksh93 supports complex relative (including English like above, crontab, at, ISO 8601:2004 durations) or absolute time specifications (see the test data in the source for some examples).

In ksh93, if you want to pass an epoch time, you use a # prefix:

$ printf '%(%c)Tn' '#0'
Thu Jan 1 00:00:00 1970

(though here, it happens to give the incorrect time for me in a British timezone where epoch time 0 was at 1am)