Manipulate date in UNIX

Question

I want to find a way to list all the dates of the current week in UNIX. For example if today is 17-07-2018 then how to get dates 16-07, 18-07, 19-07 and so on for the whole week

Answer 1

Assuming your date recognizes TZ offsets in hours, this works for the current date:

#!/bin/sh

case $(date +%A) in
  (Monday)    off="  +0  -24 -48 -72 -96 -120 -144";;
  (Tuesday)   off=" +24   +0 -24 -48 -72  -96 -120";;
  (Wednesday) off=" +48  +24  +0 -24 -48  -72  -96";;
  (Thursday)  off=" +72  +48 +24  +0 -24  -48  -72";;
  (Friday)    off=" +96  +72 +48 +24  +0  -24  -48";;
  (Saturday)  off="+120  +96 +72 +48 +24   +0  -24";;
  (Sunday)    off="+144 +120 +96 +72 +48  +24   +0";;
esac
for o in $off; do
   TZ="$o" date +%d-%m
done

Sample run:

$ date
Sun Jul 15 23:13:43 CEST 2018
$ ./x.sh
09-07
10-07
11-07
12-07
13-07
14-07
15-07

This has no problem with leap years, under/overflow, since it uses date's built-in knowledge.

PS: I just discovered that my date (on FreeBSD) takes these offsets from UTC, not the systems time zone; you should investigate this for AIX and if so, adjust the offsets accordingly by your local time zone offset to UTC.